Choose the correct option from given four options in each of the Exercises from 1 to 11. ∫ cos2x−cos2θcosx−cosθ dx is equal to (A) 2(sinx+xcosθ)+C (B) 2(sinx−xcosθ)+C (C) 2(sinx+2xcosθ)+C (D) 2(sinx−2xcosθ)+C

This is a Objective answer type Questions as classified in NCERT Exemplar Sol: Let I=∫cos2x−cos2θcosx−cosθ dx =∫(2cos2x−1)−(2cos2θ−1)cosx−cosθ dx =∫2cos2x−1−2cos2θ+1cosx−cosθ dx =∫2cos2x−2cos2θcosx−cosθ dx=2∫cos2x−cos2θcosx−cosθ dx =2∫(cosx+cosθ)(cosx−cosθ)(cosx−cosθ) dx =2∫(cosx+cosθ)dx=2(sinx+cosθ.x)+CHence, correct option is (a).

∫dxsin(x−a)sin(x−b) is equal to (A) sin(b−a)log| sin(x−a)sin(x−b) |+C (B) cosec(b−a)log| sin(x−a)sin(x−b) |+C (C) cosec(b−a)log| sin(x−a)sin(x−b) |+C (D) sin(b−a)log| sin(x−a)sin(x−b) |+C

This is a Objective answer type Questions as classified in NCERT Exemplar Sol: L e t I = ∫ d x s i n ( x − a ) . s i n ( x − b ) M u l t i p l y i n g a n d d i v i d i n g b y s i n ( b − a ) = 1 s i n ( b − a ) ∫ s i n ( b − a ) s i n ( x − a ) . s i n ( x − b ) d x = 1 s i n ( b − a ) ∫ s i n ( x + b − x − a ) s i n ( x − a ) . s i n ( x − b ) d x = 1 s i n ( b − a ) ∫ s i n [ ( x − a ) − ( x − b ) ] s i n ( x − a ) . s i n ( x − b ) d x = 1 s i n ( b − a ) ∫ s i n [ ( x − a ) − ( x − b ) ] s i n ( x − a ) . s i n ( x − b ) d x = 1 s i n ( b − a ) ∫ s i n ( x − a ) c o s ( x − b ) − c o s ( x − a ) s i n ( x − b ) s i n ( x − a ) . s i n ( x − b ) d x = 1 s i n ( b − a ) ∫ s i n ( x − a ) c o s ( x − b ) s i n ( x − a ) . s i n ( x − b ) − c o s ( x − a ) s i n ( x − b ) s i n ( x − a ) . s i n ( x − b ) d x = 1 s i n ( b − a ) ∫ [ c o s ( x − b ) s i n ( x − b ) − c o s ( x − a ) s i n ( x − a ) ] d x = 1 s i n ( b − a ) ∫ [ c o t ( x − b ) − c o t ( x − a ) ] d x = 1 s i n ( b − a ) [ l o g s i n ( x − b ) − l o g s i n ( x − a ) ] + C = 1 s i n ( b − a ) . l o g | s i n ( x − b ) s i n ( x − a ) | + C I = c o s e c ( b − a ) . l o g | s i n ( x − b ) s i n ( x − a ) | + C H e n c e , c o r r e c t o p t i o n i s ( c ) .

∫ex(1−x1+x2)2 dx (A) ex1+x2 + C (B) −ex1+x2+ C (C) ex(1+x2)2 + C (D) −ex(1+x2)2 + C

This is a Objective answer type Questions as classified in NCERT Exemplar Sol: L e t I = ∫ e x ( 1 − x 1 + x ) 2 d x = ∫ e x [ 1 + x 2 − 2 x ( 1 + x 2 ) 2 ] d x = ∫ e x [ 1 + x 2 ( 1 + x 2 ) 2 − 2 x ( 1 + x 2 ) 2 ] d x = ∫ e x [ 1 1 + x 2 − 2 x ( 1 + x 2 ) 2 ] d x H e r e f ( x ) = 1 1 + x 2 ∴ f ' ( x ) = − 2 x ( 1 + x 2 ) 2 Using ∫ex[f(x)+f'(x)]dx=ex.f(x)+C ∴ I = e x . 1 1 + x 2 + C = e x 1 + x 2 + C H e n c e , c o r r e c t o p t i o n i s ( a ) .

∫x9(4x2+1)6 dx is equal to (A) 15x(4+1x2)−5+C (B) 15(4+1x2)−5+C (C) 110(1+4)−5+C (D) 110(1x2+4)−5+C

This is a Objective answer type Questions as classified in NCERT Exemplar Sol: L e t I = ∫ x 9 ( 4 x 2 + 1 ) 6 d x = ∫ x 9 x 1 2 ( 4 + 1 x 2 ) 6 d x = ∫ 1 x 3 ( 4 + 1 x 2 ) 6 d x P u t ( 4 + 1 x 2 ) = t ⇒ − 2 x 3 d x = d t ⇒ d x x 3 = − 1 2 d t ∴ I = − 1 2 ∫ d t t 6 = − 1 2 × − 1 5 t − 5 + C = 1 1 0 ( 4 + 1 x 2 ) − 5 + C H e n c e , c o r r e c t o p t i o n i s ( d ) .

∫etan−1x (1+x+x21+x2)dx

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: Let I=∫etan−1x(1+x+x21+x2) dxPut tan−1x=t ⇒11+x2.dx=dt =∫et(1+tant+tan2t) dt=∫et(sec2t+tant) dtHere, f(t)=tant∴ f'(t)=sec2t =et.f(t)=ettant=etan−1x.x+C [? ∫et[f(x)+f'(x)] dx=exf(x)+C]Hence, I=etan−1x.x+C

Kindly consider the following ∫π2cosx esin x dx is equal to _______.

This is a Fill in the Blanks type Questions as classified in NCERT Exemplar Sol: Let I=∫0π2cosx.esinxdxPut sinx=t ⇒cosx dx=dtWhen x=0 then t=sin0=0 When x=π2 then t=sinπ2=1∴ I=∫01et dt= [et]01= (e1−e0)=e−1Hence, I=e−1

Kindly consider the following ∫x+3(x+4)2ex dx =_______.

This is a Fill in the Blanks type Questions as classified in NCERT Exemplar Sol: L e t I = ∫ x + 3 ( x + 4 ) 2 . e x d x = ∫ x + 4 − 1 ( x + 4 ) 2 . e x d x = ∫ [ x + 4 ( x + 4 ) 2 − 1 ( x + 4 ) 2 ] . e x d x = ∫ [ 1 x + 4 − 1 ( x + 4 ) 2 ] . e x d x P u t 1 x + 4 = t ⇒ − 1 ( x + 4 ) 2 d x = d t L e t f ( x ) = 1 x + 4 ∴ f ' ( x ) = − 1 ( x + 4 ) 2 Using ∫ex[f(x)+f'(x)]dx=exf(x)+C ∴ I = e x . 1 x + 4 + C H e n c e , I = e x x + 4 + C .

Kindly consider the following If: ∫a11+4x2 dx=π8 ,then, a= _______.

This is a Fill in the Blanks type Questions as classified in NCERT Exemplar Sol: L e t I = ∫ 0 a 1 1 + 4 x 2 d x = π 8 ⇒ 1 4 ∫ 0 a 1 ( 1 4 + x 2 ) d x = π 8 ⇒ ∫ 0 a 1 [ ( 1 2 ) 2 + x 2 ] d x = π 2 ⇒ 1 1 / 2 [ t a n − 1 x 1 / 2 ] 0 a = π 2 ⇒ 2 [ t a n − 1 2 a − t a n − 1 0 ] = π 2 ⇒ t a n − 1 2 a = π 4 ⇒ 2 a = t a n π 4 ⇒ 2 a = 1 ⇒ a = 1 2 H e n c e , t h e v a l u e o f a = 1 2 .

Kindly consider the following ∫2x−12x+3 dx=x−log?(2x+3)2?+C

This is a Short answer type Questions as classified in NCERT Exemplar Sol: L.H.S.=∫2x−12x+3 dx ⇒∫(1−42x+3) dx [Dividing the numerator by the denominator ] ⇒∫1. dx−4∫12x+3 dx ⇒∫1. dx−42∫1x+32 dx ⇒∫1. dx−2∫1x+32 dx ⇒x−2log|x+32|+C ⇒x−2log|2x+32|+C ⇒x−log|(2x+32)2|+C [?nlogm=logmn] ⇒x−log|(2x+3)2|−log22+C ⇒x−log|(2x+3)2|+C1=R.H.S. [where C1=C−log22]L.H.S.=R.H.S.Hence, proved.

Kindly consider the following ∫2x+3x2+3x dx=log∣x2+3x∣+C

This is a Short answer type Questions as classified in NCERT Exemplar Sol: L . H . S . = ∫ 2 x + 3 x 2 + 3 x d x P u t x 2 + 3 x = t ∴ ( 2 x + 3 ) d x = d t ⇒ ∫ d t t = l o g | t | ⇒ l o g | x 2 + 3 x | + C = R . H . S . L . H . S . = R . H . S . H e n c e , p r o v e d .

Evaluate the following: ∫(x2+2)x+1 dx

This is a Short answer type Questions as classified in NCERT Exemplar Sol: L e t I = ∫ ( x 2 + 2 ) x + 1 d x ∴ I = ∫ [ ( x − 1 ) + 3 x + 1 ] d x = ∫ ( x − 1 ) d x + 3 ∫ 1 x + 1 d x = x 2 2 − x + 3 l o g | x + 1 | + C H e n c e , t h e r e q u i r e d s o l u t i o n i s x 2 2 − x + 3 l o g | x + 1 | + C .

Kindly consider the following ∫(e6logxe4logx−e5logxe3logx) dx

This is a Short answer type Questions as classified in NCERT Exemplar Sol: L e t I = ∫ e 6 l o g x − e 5 l o g x e 4 l o g x − e 3 l o g x d x ∴ I = ∫ e l o g x 6 − e l o g x 5 e l o g x 4 − e l o g x 3 d x = ∫ x 6 − x 5 x 4 − x 3 d x = ∫ x 2 ( x 4 − x 3 ) x 4 − x 3 d x = ∫ x 2 d x = 1 3 x 3 + C H e n c e , t h e r e q u i r e d s o l u t i o n i s 1 3 x 3 + C .

Kindly consider the following ∫1+cosxx+sinx dx

This is a Short answer type Questions as classified in NCERT Exemplar Sol: L e t I = ∫ 1 + c o s x x + s i n x d x P u t x + s i n x = t ⇒ ( 1 + c o s x ) d x = d t ∴ I = ∫ d t t = l o g | t | = l o g | x + s i n x | + C H e n c e , t h e r e q u i r e d s o l u t i o n i s l o g | x + s i n x | + C .

Maths NCERT Exemplar Solutions Class 12th Chapter Seven: Overview, Questions, Preparation

Updated on Apr 11, 2025 10:03 IST

By Vishal Baghel, Executive Content Operations

Table of content

Integrals Long Answer Type Questions
Integrals Short Answer Type Questions
Integrals Objective Type Questions
Integrals Fill in the Blanks

Integrals Long Answer Type Questions

Q1. $\int^{} \frac{x^{2} d x}{x^{4} - x^{2} - 12}$

Q2. $\int \frac{x^{2} d x}{(x^{2} + a^{2}) (x^{2} + b^{2})}$

Sol:

$\begin{array}{l} L e t I = \int \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x \\ P u t x^{2} = t f o r t h e p u r p o s e o f p a r t i a l f r a c t i o n . \\ W e g e t \frac{t}{(t + a^{2}) (t + b^{2})} \\ P u t \frac{t}{(t + a^{2}) (t + b^{2})} = \frac{A}{t + a^{2}} + \frac{B}{t + b^{2}} [w h e r e A a n d B a r e a r b i t r a r yconstants .] \\ \frac{t}{(t + a^{2}) (t + b^{2})} = \frac{A (t + b^{2}) + B (t + a^{2})}{(t + a^{2}) (t + b^{2})} \\ \Rightarrow t = A t + A b^{2} + B t + B a^{2} \\ C o m p a r i n g t h e l i k e t e r m s, w e g e t \\ A + B = 1 a n d A b^{2} + B a^{2} = 0 \\ \Rightarrow A = \frac{- a^{2}}{b^{2}} B \\ ∴ \frac{- a^{2}}{b^{2}} B + B = 1 \\ B (\frac{- a^{2}}{b^{2}} + 1) = 1 \Rightarrow B (\frac{- a^{2} + b^{2}}{b^{2}}) = 1 \\ \Rightarrow B = \frac{b^{2}}{b^{2} - a^{2}} a n d A = \frac{- a^{2}}{b^{2}} \times \frac{b^{2}}{b^{2} - a^{2}} = \frac{a^{2}}{a^{2} - b^{2}} \\ S o, A = \frac{a^{2}}{a^{2} - b^{2}} a n d B = \frac{- b^{2}}{a^{2} - b^{2}} \\ ∴ \int \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x = \frac{a^{2}}{a^{2} - b^{2}} \int \frac{1}{x^{2} + a^{2}} d x - \frac{b^{2}}{a^{2} - b^{2}} \int \frac{1}{x^{2} + b^{2}} d x \\ = \frac{a^{2}}{a^{2} - b^{2}} \times \frac{1}{a} t a n^{- 1} \frac{x}{a} - \frac{b^{2}}{a^{2} - b^{2}} . \frac{1}{b} . t a n^{- 1} \frac{x}{b} \\ = \frac{a}{a^{2} - b^{2}} t a n^{- 1} \frac{x}{a} - \frac{b}{a^{2} - b^{2}} t a n^{- 1} \frac{x}{b} + C \\ H e n c e, I = \frac{1}{a^{2} - b^{2}} [a t a n^{- 1} \frac{x}{a} - b t a n^{- 1} \frac{x}{b}] + C . \end{array}$

Commonly asked questions

Kindly consider the following

$\int^{} \frac{x^{2} d x}{x^{4} - x^{2} - 12}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

$\int \frac{x^{2} d x}{(x^{2} + a^{2}) (x^{2} + b^{2})}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

This is a Long Answer Type Questions as classified in NCERT Exemplar

$\int \frac{2 x - 1}{(x - 1) (x - 2) (x - 3)} d x$

This is a Long Answer Type Questions as classified in NCERT Exemplar

$\int e^{t a n^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int e^{t a n^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x \\ P u t t a n^{- 1} x = t \Rightarrow \frac{1}{1 + x^{2}} . d x = d t \\ = \int e^{t} (1 + t a n t + t a n^{2} t) d t = \int e^{t} (s e c^{2} t + t a n t) d t \\ H e r e, f (t) = t a n t \\ ∴ f^{'} (t) = s e c^{2} t \\ = e^{t} . f (t) = e^{t} t a n t = e^{t a n^{- 1} x} . x + C \\ [? \int e^{t} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C] \\ H e n c e, I = e^{t a n^{- 1} x} . x + C \end{array}$

Kindly consider the following

This is a Long Answer Type Questions as classified in NCERT Exemplar

Kindly consider the following

This is a Long Answer Type Questions as classified in NCERT Exemplar

Kindly consider the following

$\int\sqrttanx d x (Hint: Put t a n x = t^{2})$

This is a Long answer type Questions as classified in NCERT Exemplar

Kindly consider the following

This is a Long answer type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

This is a Long answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int_{0}^{1} \underset{I I}{x l o g} \underset{I}{| 1 + 2 x |} d x \\ = {[l o g | 1 + 2 x | . (\frac{x^{2}}{2})]}_{0}^{1} - \int_{0}^{1} (\frac{1.2}{1 + 2 x} . \frac{x^{2}}{2}) d x \\ = \frac{1}{2} {[x^{2} l o g (1 + 2 x)]}_{0}^{1} - \int_{0}^{1} (\frac{x^{2}}{1 + 2 x}) d x \\ = \frac{1}{2} [l o g 3 - 0] - \int_{0}^{1} (\frac{x}{2} - \frac{x / 2}{1 + 2 x}) d x \\ = \frac{1}{2} l o g 3 - \frac{1}{2} \int_{0}^{1} x d x + \frac{1}{2} \int_{0}^{1} \frac{x}{1 + 2 x} d x \\ = \frac{1}{2} l o g 3 - \frac{1}{2} {[\frac{x^{2}}{2}]}_{0}^{1} + \frac{1}{2} . \frac{1}{2} \int_{0}^{1} \frac{(2 x + 1 - 1)}{1 + 2 x} d x \\ = \frac{1}{2} l o g 3 - \frac{1}{4} [1 - 0] + \frac{1}{4} \int_{0}^{1} 1 d x - \frac{1}{4} \int_{0}^{1} \frac{1}{2 x + 1} d x \\ = \frac{1}{2} l o g 3 - \frac{1}{4} + \frac{1}{4} {[x]}_{0}^{1} - \frac{1}{4} . \frac{1}{2} {[l o g | 2 x + 1 |]}_{0}^{1} \\ = \frac{1}{2} l o g 3 - \frac{1}{4} + \frac{1}{4} - \frac{1}{8} [l o g 3 - 0] \\ = \frac{1}{2} l o g 3 - \frac{1}{8} l o g 3 = \frac{3}{8} l o g 3 \\ H e n c e, I = \frac{3}{8} l o g 3 . \end{array}$

Kindly consider the following

This is a Long answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int_{0}^{π} x l o g s i n x d x \dots (i) \\ = \int_{0}^{π} (π - x) l o g s i n (π - x) d x [\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] \\ = \int_{0}^{π} (π - x) l o g s i n x d x \dots (i i) \\ A d d i n g (i) a n d (i i) \\ 2 I = \int_{0}^{π} [(π - x) l o g s i n x + x l o g s i n x] d x \\ 2 I = \int_{0}^{π} π l o g s i n x d x \\ 2 I = 2 π \int_{0}^{\frac{π}{2}} l o g s i n x d x [? \int_{0}^{a} f (x) d x = 2 \int_{0}^{a / 2} f (x) d x] \\ ∴ I = π \int_{0}^{\frac{π}{2}} l o g s i n x d x \dots (i i i) \\ I = π \int_{0}^{\frac{π}{2}} l o g s i n (\frac{π}{2} - x) d x \\ I = π \int_{0}^{\frac{π}{2}} l o g c o s x d x \dots (i v) \\ O n a d d i n g (i i i) a n d (i v), w e g e t \\ 2 I = π \int_{0}^{\frac{π}{2}} (l o g s i n x + l o g c o s x) d x \\ 2 I = π \int_{0}^{\frac{π}{2}} l o g s i n x c o s x d x \\ 2 I = π \int_{0}^{\frac{π}{2}} \frac{l o g 2 s i n x c o s x}{2} d x \\ 2 I = π \int_{0}^{\frac{π}{2}} l o g s i n 2 x d x - π \int_{0}^{\frac{π}{2}} l o g 2 d x \\ P u t 2 x = t \Rightarrow 2 d x = d t \Rightarrow d x = \frac{d t}{2} \\ 2 I = π \int_{0}^{π} l o g s i n t d t - π . l o g 2 \int_{0}^{\frac{π}{2}} 1 d x [C h a n g i n g t h e limit] \\ 2 I = I - π . l o g 2 {[x]}_{0}^{\frac{π}{2}} & \end{array}$

Kindly consider the following

This is a Long answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int_{- \frac{π}{4}}^{\frac{π}{4}} l o g | s i n x + c o s x | d x \dots (i) \\ = \int_{- \frac{π}{4}}^{\frac{π}{4}} l o g | s i n (\frac{π}{4} - \frac{π}{4} - x) + c o s (\frac{π}{4} - \frac{π}{4} - x) | d x [Using \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] \\ = \int_{- \frac{π}{4}}^{\frac{π}{4}} l o g | s i n (- x) + c o s x | d x \\ = \int_{- \frac{π}{4}}^{\frac{π}{4}} l o g | c o s x - s i n x | d x \dots (i i) \\ A d d i n g (i) a n d (i i) \\ 2 I = \int_{- \frac{π}{4}}^{\frac{π}{4}} l o g | c o s x + s i n x | d x + \int_{- \frac{π}{4}}^{\frac{π}{4}} l o g | c o s x - s i n x | d x \\ 2 I = \int_{- \frac{π}{4}}^{\frac{π}{4}} l o g | (c o s x + s i n x) (c o s x - s i n x) | d x \\ 2 I = \int_{- \frac{π}{4}}^{\frac{π}{4}} l o g | c o s^{2} x - s i n^{2} x | d x \\ ∴ 2 I = \int_{- \frac{π}{4}}^{\frac{π}{4}} l o g c o s 2 x d x \\ 2 I = 2 \int_{0}^{\frac{π}{4}} l o g c o s 2 x d x [? \int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x i f f (- x) = f (x)] \\ ∴ I = π \int_{0}^{\frac{π}{4}} l o g c o s 2 x d x \\ P u t 2 x = t d x = \frac{d t}{2} \\ W h e n x = 0 ∴ t = 0; w h e n x = \frac{π}{4} ∴ t = \frac{π}{2} \\ I = \frac{1}{2} \int_{0}^{\frac{π}{2}} l o g c o s t d t &thi \end{array}$

$\begin{array}{l} O n a d d i n g (i i i) a n d (i v), w e g e t \\ 2 I = \frac{1}{2} \int_{0}^{\frac{π}{2}} (l o g c o s t + l o g s i n t) d t \\ 2 I = \frac{1}{2} \int_{0}^{\frac{π}{2}} l o g s i n t c o s t d t \\ 2 I = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{l o g 2 s i n t c o s t d t}{2} \\ 2 I = \frac{1}{2} \int_{0}^{\frac{π}{2}} (l o g s i n 2 t - l o g 2) d t \\ 4 I = \int_{0}^{\frac{π}{2}} l o g s i n 2 t d t - \int_{0}^{\frac{π}{2}} l o g 2 d t \\ P u t 2 t = u \Rightarrow 2 d t = d u \Rightarrow d t = \frac{d u}{2} \\ ∴ 4 I = \frac{1}{2} \int_{0}^{π} l o g s i n u d u - \int_{0}^{\frac{π}{2}} l o g 2 d t [C h a n g i n g t h e limit] \\ 4 I = \frac{1}{2} \times 2 \int_{0}^{\frac{π}{2}} l o g s i n u d u - l o g 2 {[t]}_{0}^{\frac{π}{2}} \\ 4 I = \int_{0}^{\frac{π}{2}} l o g s i n u d u - l o g 2 . \frac{π}{2} \\ 4 I = 2 I - \frac{π}{2} . l o g 2 \end{array}$

Integrals Short Answer Type Questions

Q1. $\int \frac{2 x - 1}{2 x + 3} d x = x - l o g ∣ {(2 x + 3)}^{2} ∣ + C$

Sol:

$\begin{array}{l} L . H . S . = \int \frac{2 x - 1}{2 x + 3} d x \\ \Rightarrow \int (1 - \frac{4}{2 x + 3}) d x [D i v i d i n g t h e n u m e r a t o r b y t h e denominator] \\ \Rightarrow \int 1 . d x - 4 \int \frac{1}{2 x + 3} d x \Rightarrow \int 1 . d x - \frac{4}{2} \int \frac{1}{x + \frac{3}{2}} d x \\ \Rightarrow \int 1 . d x - 2 \int \frac{1}{x + \frac{3}{2}} d x \Rightarrow x - 2 l o g | x + \frac{3}{2} | + C \\ \Rightarrow x - 2 l o g | \frac{2 x + 3}{2} | + C \Rightarrow x - l o g | {(\frac{2 x + 3}{2})}^{2} | + C \\ [∵ n l o g m = l o g m^{n}] \\ \Rightarrow x - l o g | {(2 x + 3)}^{2} | - l o g 2^{2} + C \\ \Rightarrow x - l o g | {(2 x + 3)}^{2} | + C_{1} = R . H . S . [w h e r e C_{1} = C - l o g 2^{2}] \\ L . H . S . = R . H . S . \\ H e n c e, p r o v e d . \end{array}$

Q2. $\int \frac{2 x + 3}{x^{2} + 3 x} d x = l o g ∣ x^{2} + 3 x ∣ + C$

Sol:

$\begin{array}{l} L . H . S . = \int \frac{2 x + 3}{x^{2} + 3 x} d x \\ P u t x^{2} + 3 x = t \\ ∴ (2 x + 3) d x = d t \\ \Rightarrow \int \frac{d t}{t} = l o g | t | \Rightarrow l o g | x^{2} + 3 x | + C = R . H . S . \\ L . H . S . = R . H . S . \\ H e n c e, p r o v e d . \end{array}$

Commonly asked questions

Kindly consider the following

$\int \frac{2 x - 1}{2 x + 3} d x = x - l o g ? {(2 x + 3)}^{2} ? + C$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L . H . S . = \int \frac{2 x - 1}{2 x + 3} d x \\ \Rightarrow \int (1 - \frac{4}{2 x + 3}) d x [D i v i d i n g t h e n u m e r a t o r b y t h e denominator] \\ \Rightarrow \int 1 . d x - 4 \int \frac{1}{2 x + 3} d x \Rightarrow \int 1 . d x - \frac{4}{2} \int \frac{1}{x + \frac{3}{2}} d x \\ \Rightarrow \int 1 . d x - 2 \int \frac{1}{x + \frac{3}{2}} d x \Rightarrow x - 2 l o g | x + \frac{3}{2} | + C \\ \Rightarrow x - 2 l o g | \frac{2 x + 3}{2} | + C \Rightarrow x - l o g | {(\frac{2 x + 3}{2})}^{2} | + C \\ [? n l o g m = l o g m^{n}] \\ \Rightarrow x - l o g | {(2 x + 3)}^{2} | - l o g 2^{2} + C \\ \Rightarrow x - l o g | {(2 x + 3)}^{2} | + C_{1} = R . H . S . [w h e r e C_{1} = C - l o g 2^{2}] \\ L . H . S . = R . H . S . \\ H e n c e, p r o v e d . \end{array}$

Kindly consider the following

$\int \frac{2 x + 3}{x^{2} + 3 x} d x = l o g ∣ x^{2} + 3 x ∣ + C$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

Evaluate the following:

$\int \frac{(x^{2} + 2)}{x + 1} d x$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int \frac{(x^{2} + 2)}{x + 1} d x \\ ∴ I = \int [(x - 1) + \frac{3}{x + 1}] d x \\ = \int (x - 1) d x + 3 \int \frac{1}{x + 1} d x \\ = \frac{x^{2}}{2} - x + 3 l o g | x + 1 | + C \\ H e n c e, t h e r e q u i r e d s o l u t i o n i s \frac{x^{2}}{2} - x + 3 l o g | x + 1 | + C . \end{array}$

Kindly consider the following

$\int (\frac{e^{6 l o g x}}{e^{4 l o g x}} - \frac{e^{5 l o g x}}{e^{3 l o g x}}) d x$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int \frac{e^{6 l o g x} - e^{5 l o g x}}{e^{4 l o g x} - e^{3 l o g x}} d x \\ ∴ I = \int \frac{e^{l o g x^{6}} - e^{l o g x^{5}}}{e^{l o g x^{4}} - e^{l o g x^{3}}} d x \\ = \int \frac{x^{6} - x^{5}}{x^{4} - x^{3}} d x = \int \frac{x^{2} (x^{4} - x^{3})}{x^{4} - x^{3}} d x = \int x^{2} d x \\ = \frac{1}{3} x^{3} + C \\ H e n c e, t h e r e q u i r e d s o l u t i o n i s \frac{1}{3} x^{3} + C . \end{array}$

Kindly consider the following

$\int \frac{1 + c o s x}{x + s i n x} d x$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int \frac{1 + c o s x}{x + s i n x} d x \\ P u t x + s i n x = t \Rightarrow (1 + c o s x) d x = d t \\ ∴ I = \int \frac{d t}{t} = l o g | t | = l o g | x + s i n x | + C \\ H e n c e, t h e r e q u i r e d s o l u t i o n i s l o g | x + s i n x | + C . \end{array}$

Kindly consider the following

$\int \frac{d x}{1 + c o s x}$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int \frac{d x}{1 + c o s x} \\ = \int \frac{d x}{2 c o s^{2} x / 2} [? 1 + c o s x = 2 c o s^{2} x / 2] \\ = \frac{1}{2} \int s e c^{2} \frac{x}{2} d x = \frac{1}{2} . 2 t a n \frac{x}{2} + C = t a n \frac{x}{2} + C \\ H e n c e, t h e r e q u i r e d s o l u t i o n i s t a n \frac{x}{2} + C . \end{array}$

Kindly consider the following

$\int t a n^{2} x s e c^{4} x d x$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int t a n^{2} x . s e c^{4} x d x \\ = \int t a n^{2} x . s e c^{2} x . s e c^{2} x d x = \int t a n^{2} x . (1 + t a n^{2} x) . s e c^{2} x d x \\ P u t t a n x = t, ∴ s e c^{2} x d x = d t \\ ∴ I = \int t^{2} (1 + t^{2}) d t = \int (t^{2} + t^{4}) d t = \int t^{2} d t + \int t^{4} d t \\ = \frac{1}{3} t^{3} + \frac{1}{5} t^{5} = \frac{1}{3} t a n^{3} x + \frac{1}{5} t a n^{5} x + C \\ H e n c e, t h e r e q u i r e d s o l u t i o n i s \frac{1}{3} t a n^{3} x + \frac{1}{5} t a n^{5} x + C . \end{array}$

Kindly consider the following

This is a Short answer type Questions as classified in NCERT Exemplar

Kindly consider the following

$\int √1+sinx d x$

This is a Short answer type Questions as classified in NCERT Exemplar

Kindly consider the following

This is a Short answer type Questions as classified in NCERT Exemplar

Kindly consider the following

This is a Short answer type Questions as classified in NCERT Exemplar

Kindly consider the following

$\int \frac{x^{\frac{1}{2}}}{1 + x^{\frac{3}{4}}} d x (x = z^{4})$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int \frac{x^{1 / 2}}{1 + x^{3 / 4}} d x \\ P u t x = t^{4} \Rightarrow d x = 4 t^{3} d t \\ = \int \frac{t^{2} . 4 t^{3}}{1 + t^{3}} d t = 4 \int \frac{t^{5}}{1 + t^{3}} d t \\ = 4 \int (t^{2} - \frac{t^{2}}{t^{3} + 1}) d t = 4 \int t^{2} d t - 4 \int \frac{t^{2}}{t^{3} + 1} d t \\ L e t I = I_{1} - I_{2} \\ N o w I_{1} = 4 \int t^{2} d t = 4 . \frac{t^{3}}{3} + C_{1} = \frac{4}{3} x^{3 / 4} + C_{1} \\ a n d I_{2} = 4 \int \frac{t^{2}}{t^{3} + 1} d t \\ P u t t^{3} + 1 = z \Rightarrow 3 t^{2} d t = d z \Rightarrow t^{2} d t = \frac{1}{3} d z \\ ∴ I_{2} = \frac{4}{3} \int \frac{d z}{z} = \frac{4}{3} l o g | z | + C_{2} = \frac{4}{3} l o g | t^{3} + 1 | + C_{2} \\ = \frac{4}{3} l o g | {(x)}^{3 / 4} + 1 | + C_{2} \\ ∴ I = I_{1} - I_{2} \\ = \frac{4}{3} x^{3 / 4} + C_{1} - \frac{4}{3} l o g | {(x)}^{3 / 4} + 1 | - C_{2} \\ ∴ I = \frac{4}{3} x^{3 / 4} - \frac{4}{3} l o g | {(x)}^{3 / 4} + 1 | + C_{1} - C_{2} \\ H e n c e, I = \frac{4}{3} [x^{3 / 4} - l o g | {(x)}^{3 / 4} + 1 |] + C . [? C_{1} - C_{2} = C] \end{array}$

Kindly consider the following

This is a Short answer type Questions as classified in NCERT Exemplar

Kindly consider the following

This is a Short answer type Questions as classified in NCERT Exemplar

Kindly consider the following

This is a Short answer type Questions as classified in NCERT Exemplar

Kindly consider the following

This is a Short answer type Questions as classified in NCERT Exemplar

Kindly consider the following

This is a Short answer type Questions as classified in NCERT Exemplar

Kindly consider the following

$\int \frac{x}{x^{4} - 1} d x$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int \frac{x}{x^{4} - 1} d x \\ P u t x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} \\ = \frac{1}{2} \int \frac{d t}{t^{2} - 1} = \frac{1}{2} \int \frac{d t}{t^{2} - {(1)}^{2}} = \frac{1}{2} . \frac{1}{2.1} l o g | \frac{t - 1}{t + 1} | + C \\ [? \frac{1}{2} \int \frac{d t}{x^{2} - a^{2}} = \frac{1}{2 a} l o g | \frac{x - a}{x + a} | + C] \\ = \frac{1}{4} l o g | \frac{x^{2} - 1}{x^{2} + 1} | + C \\ H e n c e, I = \frac{1}{4} l o g | \frac{x^{2} - 1}{x^{2} + 1} | + C . \end{array}$

Kindly consider the following

$\int \frac{x^{2}}{1 - x^{4}} d x (x^{2} = t)$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int \frac{x^{2}}{1 - x^{4}} d x \\ = \int \frac{x^{2}}{(1 - x^{2}) (1 + x^{2})} d x \\ P u t x^{2} = t f o r t h e p u r p o s e o f p a r t i a l f r a c t i o n s . \\ W e g e t \frac{t}{(1 - t) (1 + t)} \\ Resolvinginto p a r t i a l f r a c t i o n s w e p u t \\ \frac{t}{(1 - t) (1 + t)} = \frac{A}{1 - t} + \frac{B}{1 + t} [W h e r e A a n d B a r e a r b i t a r y] \\ \Rightarrow \frac{t}{(1 - t) (1 + t)} = \frac{A (1 + t) + B (1 - t)}{(1 - t) (1 + t)} \\ \Rightarrow t = A + A t + B - B t \\ C o m p a r i n g t h e l i k e t e r m s, w e g e t A - B = 1 a n d A + B = 0 \\ S o l v i n g t h e a b o v e e q u a t i o n s, w e h a v e A = \frac{1}{2} a n d B = - \frac{1}{2} \\ ∴ I = \int \frac{\frac{1}{2}}{1 - x^{2}} d x + \int \frac{- \frac{1}{2}}{1 + x^{2}} d x [P u t t i n g t = x^{2}] \\ = \frac{1}{2} . \frac{1}{2.1} l o g | \frac{1 + x}{1 - x} | - \frac{1}{2} t a n^{- 1} x + C \\ = \frac{1}{4} l o g | \frac{1 + x}{1 - x} | - \frac{1}{2} t a n^{- 1} x + C \\ H e n c e, I = \frac{1}{4} l o g | \frac{1 + x}{1 - x} | - \frac{1}{2} t a n^{- 1} x + C . \end{array}$

Kindly consider the following

This is a Short answer type Questions as classified in NCERT Exemplar

Kindly consider the following

$\int \frac{s i n^{- 1}}{{(1 - x^{2})}^{\frac{3}{2}}} d x$

This is a Short answer type Questions as classified in NCERT Exemplar

Kindly consider the following

$\int \frac{(c o s 5 x + c o s 4 x}{1 - 2 c o s 3 x} d x$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int \frac{c o s 5 x + c o s 4 x}{1 - 2 c o s 3 x} d x \\ I = \int \frac{2 c o s \frac{5 x + 4 x}{2} . c o s \frac{5 x - 4 x}{2}}{1 - 2 (c o s^{2} \frac{3 x}{2} - 1)} d x \\ = \int \frac{2 c o s \frac{9 x}{2} . c o s \frac{x}{2}}{1 - 4 c o s^{2} \frac{3 x}{2} + 2} d x = \int \frac{2 c o s \frac{9 x}{2} . c o s \frac{x}{2}}{3 - 4 c o s^{2} \frac{3 x}{2}} d x \\ = - \int \frac{2 c o s \frac{9 x}{2} . c o s \frac{x}{2}}{4 c o s^{2} \frac{3 x}{2} - 3} d x = - \int \frac{2 c o s \frac{9 x}{2} . c o s \frac{x}{2} . c o s \frac{3 x}{2}}{4 c o s^{3} \frac{3 x}{2} - 3 c o s \frac{3 x}{2}} d x [\begin{array}{l} M u l t i p l y i n g a n d d i v i d i n g \\ b y c o s \frac{3 x}{2} \end{array}] \\ = - \int \frac{2 c o s \frac{9 x}{2} . c o s \frac{x}{2} . c o s \frac{3 x}{2}}{c o s 3 . \frac{3 x}{2}} d x [? c o s 3 x = 4 c o s^{3} x - 3 c o s x] \\ = - \int \frac{2 c o s \frac{9 x}{2} . c o s \frac{x}{2} . c o s \frac{3 x}{2}}{c o s \frac{9 x}{2}} d x = - \int 2 c o s \frac{3 x}{2} . c o s \frac{x}{2} d x \\ = - \int [c o s (\frac{3 x}{2} + \frac{x}{2}) + c o s (\frac{3 x}{2} - \frac{x}{2})] d x \\ = - \int (c o s 2 x + c o s x) d x [? 2 c o s A c o s B = c o s (A + B) + c o s (A - B)] \\ = - \int c o s 2 x d x - \int c o s x d x = - \frac{1}{2} s i n 2 x - s i n x + C \\ H e n c e, I = - [\frac{1}{2} s i n 2 x + s i n x] + C \end{array}$

Kindly consider the following

$\int \frac{s i n^{6} x + c o s^{6} x}{s i n^{2} x + c o s^{2} x} d x$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int \frac{s i n^{6} x + c o s^{6} x}{s i n^{2} x . c o s^{2} x} d x \\ I = \int \frac{{(s i n^{2} x + c o s^{2} x)}^{3} - 3 s i n^{2} x c o s^{2} x (s i n^{2} x + c o s^{2} x)}{s i n^{2} x . c o s^{2} x} d x \\ [? a^{3} + b^{3} = {(a + b)}^{3} - 3 a b (a + b)] \\ = \int \frac{{(1)}^{3} - 3 s i n^{2} x c o s^{2} x . (1)}{s i n^{2} x . c o s^{2} x} d x \\ = \int \frac{1 - 3 s i n^{2} x c o s^{2} x}{s i n^{2} x . c o s^{2} x} d x \\ = \int (\frac{1}{s i n^{2} x . c o s^{2} x} - \frac{3 s i n^{2} x c o s^{2} x}{s i n^{2} x . c o s^{2} x}) d x \\ = \int (\frac{1}{s i n^{2} x . c o s^{2} x} - 3) d x = \int (\frac{s i n^{2} x + c o s^{2} x}{s i n^{2} x . c o s^{2} x} - 3) d x \\ = \int [(\frac{1}{c o s^{2} x} + \frac{1}{s i n^{2} x}) - 3] d x \\ = \int (s e c^{2} x + c o s e c^{2} x - 3) d x \\ = \int s e c^{2} x d x + \int c o s e c^{2} x d x - 3 \int 1 d x \\ = t a n x - c o t x - 3 x + C \\ H e n c e, I = t a n x - c o t x - 3 x + C . \end{array}$

Kindly consider the following

This is a Short answer type Questions as classified in NCERT Exemplar

Kindly consider the following

$\int \frac{(c o s x - c o s 2 x)}{1 - c o s x} d x$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int \frac{c o s x - c o s 2 x}{1 - c o s x} d x \\ I = \int \frac{2 s i n \frac{x + 2 x}{2} . s i n \frac{2 x - x}{2}}{2 s i n^{2} \frac{x}{2}} d x [? c o s C - c o s D = 2 s i n \frac{C + D}{2} . s i n \frac{D - C}{2}] \\ = \int \frac{2 s i n \frac{3 x}{2} . s i n \frac{x}{2}}{2 s i n^{2} \frac{x}{2}} d x = \int \frac{s i n \frac{3 x}{2}}{s i n \frac{x}{2}} d x = \int \frac{s i n 3 (\frac{x}{2})}{s i n (\frac{x}{2})} d x \\ = \int \frac{3 s i n \frac{x}{2} - 4 s i n^{3} \frac{x}{2}}{s i n \frac{x}{2}} d x [s i n 3 x = 3 s i n x - 4 s i n^{3} x] \\ = \int \frac{s i n \frac{x}{2} (3 - 4 s i n^{2} \frac{x}{2})}{s i n \frac{x}{2}} d x = \int (3 - 4 s i n^{2} \frac{x}{2}) d x \\ = \int [3 - 2 (1 - c o s x)] d x [? 2 s i n^{2} \frac{x}{2} = 1 - c o s x] \\ = \int (3 - 2 + 2 c o s x) d x = \int (1 + 2 c o s x) d x \\ = x + 2 s i n x + C \\ H e n c e, I = x + 2 s i n x + C . \end{array}$

Kindly consider the following

This is a Short answer type Questions as classified in NCERT Exemplar

Evaluate the following as the limit of sums

$\int^{2} (x^{2} + 3) d x$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int_{0}^{2} (x^{2} + 3) d x \\ Using t h e f o r m u l a, \\ \int_{a}^{b} f (x) d x = \underset{h \to 0}{l i m} h [f (a) + f (a + h) + f (a + 2 h) + \dots + f (a + \bar{n - 1} h)] \\ w h e r e h = \frac{b - a}{n} H e r e, a = 0 a n d b = 2 \\ ∴ h = \frac{2 - 0}{n} ∴ n h = 2 \\ H e r e, f (x) = x^{2} + 3 \\ f (0) = 0^{2} + 3 = 0 + 3 = 3 \\ f (0 + h) = {(0 + h)}^{2} + 3 = h^{2} + 3 \\ f (0 + 2 h) = {(0 + 2 h)}^{2} + 3 = 4 h^{2} + 3 = 3 \\ \dots \dots \dots \dots \dots \dots \\ \dots \dots \dots \dots \dots \dots \\ f (0 + \bar{n - 1} h) = {(0 + \bar{n - 1} h)}^{2} + 3 {(n - 1)}^{2} h^{2} + 3 \\ N o w, \int_{0}^{2} (x^{2} + 3) d x = \underset{h \to 0}{l i m} h [3 + h^{2} + 3 + 4 h^{2} + 3 + \dots + {(n - 1)}^{2} h^{2} + 3] \\ = \underset{h \to 0}{l i m} h [(3 + 3 + 3 + \dots + n) + {h^{2} + 4 h^{2} + \dots + {(n - 1)}^{2} h^{2}}] \\ = \underset{h \to 0}{l i m} h [3 n + h^{2} {1 + 4 + \dots + {(n - 1)}^{2}}] \\ = \underset{h \to 0}{l i m} h [3 n + h^{2} \frac{n (n - 1) (2 n - 1)}{6}] \\ [? 1 + 4 + 9 + \dots + {(n - 1)}^{2} = \frac{n (n - 1) (2 n - 1)}{6}] \\ = \underset{h \to 0}{l i m} [3 n h + \frac{h^{3} n (n - 1) (2 n - 1)}{6}] \\ = \underset{h \to 0}{l i m} [3 n h + \frac{n h (n h - h) (2 n h - h)}{6}] \\ = [3 \times 2 + \frac{2 (2 - 0) (2 \times 2 - 0)}{6}] [? n h = 2 a n d h = 0] \\ = [6 + \frac{2 \times 2 \times 4}{6}] = 6 + \frac{8}{3} = \frac{26}{3} \\ H e n c e, \int_{0}^{2} (x^{2} + 3) d x = 26 \end{array}$

Kindly consider the following

$\int^{2} e^{x} d x$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int_{0}^{2} e^{x} d x \\ Using t h e f o r m u l a, \\ \int_{a}^{b} f (x) d x = \underset{h \to 0}{l i m} h [f (a) + f (a + h) + f (a + 2 h) + \dots + f (a + \bar{n - 1} h)] \\ w h e r e h = \frac{b - a}{n} H e r e, a = 0 a n d b = 2 \\ ∴ h = \frac{2 - 0}{n} ∴ n h = 2 \\ H e r e, f (x) = e^{x} \\ f (0) = e^{0} = 1 \\ f (0 + h) = e^{0 + h} = e^{h} \\ f (0 + 2 h) = e^{0 + 2 h} = e^{2 h} \\ \dots \dots \dots \dots \dots \dots \\ \dots \dots \dots \dots \dots \dots \\ f (0 + \bar{n - 1} h) = e^{0 + (n - 1) h} = e^{(n - 1) h} \\ N o w, \int_{0}^{2} e^{x} d x = \underset{h \to 0}{l i m} h [1 + e^{h} + e^{2 h} + \dots + e^{(n - 1) h}] \\ = \underset{h \to 0}{l i m} h [\frac{1 (e^{n h} - 1)}{e^{h} - 1}] [? a + a r + a r^{2} + \dots + a r^{n - 1} = \frac{a (r^{n} - 1)}{r - 1}] \\ = \underset{h \to 0}{l i m} \frac{e^{n h} - 1}{\frac{e^{h} - 1}{h}} = \frac{e^{2} - 1}{1} = e^{2} - 1 [? \underset{x \to 0}{l i m} \frac{e^{x} - 1}{x} = 1] \\ H e n c e, \int_{0}^{2} e^{2} d x = e^{2} - 1 . \end{array}$

Evaluate the following

$\int^{1} (\frac{d x}{e^{x} + e^{- x}}) d x$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int_{0}^{1} \frac{d x}{e^{x} + e^{- x}} \\ = \int_{0}^{1} \frac{d x}{e^{x} + \frac{1}{e^{x}}} = \int_{0}^{1} \frac{d x}{\frac{e^{2 x} + 1}{e^{x}}} = \int_{0}^{1} \frac{e^{x} d x}{e^{2 x} + 1} \\ P u t e^{x} = t \Rightarrow e^{x} d x = d t \\ C h a n g i n g t h e l i m i t s, w e h a v e \\ W h e n x = 0 ∴ t = e^{0} = 1 \\ W h e n x = 1 ∴ t = e^{1} = e \\ ∴ I = \int_{1}^{e} \frac{d t}{t^{2} + 1} = {[t a n^{- 1} t]}_{1}^{e} = [t a n^{- 1} e - t a n^{- 1} (1)] = t a n^{- 1} e - \frac{π}{4} \\ H e n c e, I = t a n^{- 1} e - \frac{π}{4} . \end{array}$

Kindly consider the following

$\int^{\frac{π}{2}} (\frac{t a n x d x}{1 + m^{2} t a n^{2} x})$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int_{0}^{\frac{π}{2}} \frac{t a n x}{1 + m^{2} t a n^{2} x} d x \\ I = \int_{0}^{\frac{π}{2}} \frac{\frac{s i n x}{c o s x}}{1 + m^{2} \frac{s i n^{2} x}{c o s^{2} x}} d x = \int_{0}^{\frac{π}{2}} \frac{\frac{s i n x}{c o s x}}{\frac{c o s^{2} x + m^{2} s i n^{2} x}{c o s^{2} x}} d x \\ = \int_{0}^{\frac{π}{2}} \frac{s i n x c o s x}{c o s^{2} x + m^{2} s i n^{2} x} d x = \int_{0}^{\frac{π}{2}} \frac{s i n x c o s x}{1 - s i n^{2} x + m^{2} s i n^{2} x} d x \\ = \int_{0}^{\frac{π}{2}} \frac{s i n x c o s x}{1 - s i n^{2} x + (1 - m^{2})} d x \\ P u t s i n^{2} x = t \\ 2 s i n x c o s x d x = d t \\ s i n x c o s x d x = \frac{d t}{2} \\ C h a n g i n g t h e limits w e g e t, \\ W h e n x = 0 ∴ t = s i n^{2} 0 = 0; w h e n x = \frac{π}{2} ∴ t = s i n^{2} \frac{π}{2} = 1 \\ ∴ I = \frac{1}{2} \int_{0}^{1} \frac{d t}{1 + (m^{2} - 1) t} = \frac{1}{2} {[\frac{l o g [1 + (m^{2} - 1) t]}{m^{2} - 1}]}_{0}^{1} \\ = \frac{1}{2 (m^{2} - 1)} [l o g (1 + m^{2} - 1) - l o g (1)] = \frac{l o g | m^{2} |}{2 (m^{2} - 1)} \\ H e n c e, I = \frac{l o g | m^{2} |}{2 (m^{2} - 1)} = \frac{l o g | m |}{m^{2} - 1} . \end{array}$

Kindly consider the following

This is a Short answer type Questions as classified in NCERT Exemplar

Kindly consider the following

This is a Short answer type Questions as classified in NCERT Exemplar

Kindly consider the following

$\int^{π} x$ sin x cos²xdx

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int_{0}^{π} x s i n x c o s^{2} x d x \dots (i) \\ = \int_{0}^{π} (π - x) s i n (π - x) c o s^{2} (π - x) d x \\ = \int_{0}^{π} (π - x) s i n x c o s^{2} x d x \dots (i i) \\ A d d i n g (i) a n d (i i) w e g e t, \\ 2 I = \int_{0}^{π} [x s i n x c o s^{2} x + (π - x) s i n x c o s^{2} x] d x \\ 2 I = \int_{0}^{π} [s i n x c o s^{2} x . (x + π - x)] d x \\ 2 I = \int_{0}^{π} π s i n x c o s^{2} x d x = π \int_{0}^{π} s i n x c o s^{2} x d x \\ P u t c o s x = t \Rightarrow - s i n x d x = d t \Rightarrow s i n x d x = - d t \\ C h a n g i n g t h e limits w e g e t, \\ W h e n x = 0, t = c o s 0 = 1 \\ w h e n x = π, t = c o s π = - 1 \\ ∴ 2 I = π \int_{1}^{- 1} - t^{2} d t = - π \int_{1}^{- 1} t^{2} d t \\ 2 I = π \int_{- 1}^{1} t^{2} d t [? \int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x] \\ 2 I = π {[\frac{t^{3}}{3}]}_{- 1}^{1} = π [\frac{1}{3} + \frac{1}{3}] = π (\frac{2}{3}) \\ ∴ I = \frac{π}{3} \end{array}$

Kindly consider the following

This is a Short answer type Questions as classified in NCERT Exemplar

Integrals Objective Type Questions

Choose the correct option from given four options in each of the Exercises from 1 to 11.

Q1. $\int \frac{c o s 2 x - c o s 2 θ}{c o s x - c o s θ} d x$ is equal to

(A) $2 (s i n x + x c o s θ) + C$

(B) $2 (s i n x - x c o s θ) + C$

(D) $2 (s i n x - 2 x c o s θ) + C$

Sol:

$\begin{array}{l} L e t I = \int \frac{c o s 2 x - c o s 2 θ}{c o s x - c o s θ} d x \\ = \int \frac{(2 c o s^{2} x - 1) - (2 c o s^{2} θ - 1)}{c o s x - c o s θ} d x \\ = \int \frac{2 c o s^{2} x - 1 - 2 c o s^{2} θ + 1}{c o s x - c o s θ} d x \\ = \int \frac{2 c o s^{2} x - 2 c o s^{2} θ}{c o s x - c o s θ} d x = 2 \int \frac{c o s^{2} x - c o s^{2} θ}{c o s x - c o s θ} d x \\ = 2 \int \frac{(c o s x + c o s θ) (c o s x - c o s θ)}{(c o s x - c o s θ)} d x \\ = 2 \int (c o s x + c o s θ) d x = 2 (s i n x + c o s θ . x) + C \\ H e n c e, c o r r e c t o p t i o n i s (a) . \end{array}$

Q2. $\int \frac{d x}{s i n (x - a) s i n (x - b)} i s e q u a l t o$

(A) $s i n (b - a) l o g | \frac{s i n (x - a)}{s i n (x - b)} | + C$

(B) $c o s e c (b - a) l o g | \frac{s i n (x - a)}{s i n (x - b)} | + C$

(D) $s i n (b - a) l o g | \frac{s i n (x - a)}{s i n (x - b)} | + C$

Sol:

$\begin{array}{l} L e t I = \int \frac{d x}{s i n (x - a) . s i n (x - b)} \\ M u l t i p l y i n g a n d d i v i d i n g b y s i n (b - a) \\ = \frac{1}{s i n (b - a)} \int \frac{s i n (b - a)}{s i n (x - a) . s i n (x - b)} d x \\ = \frac{1}{s i n (b - a)} \int \frac{s i n (x + b - x - a)}{s i n (x - a) . s i n (x - b)} d x \\ = \frac{1}{s i n (b - a)} \int \frac{s i n [(x - a) - (x - b)]}{s i n (x - a) . s i n (x - b)} d x \\ = \frac{1}{s i n (b - a)} \int \frac{s i n [(x - a) - (x - b)]}{s i n (x - a) . s i n (x - b)} d x \\ = \frac{1}{s i n (b - a)} \int \frac{s i n (x - a) c o s (x - b) - c o s (x - a) s i n (x - b)}{s i n (x - a) . s i n (x - b)} d x \\ = \frac{1}{s i n (b - a)} \int \frac{s i n (x - a) c o s (x - b)}{s i n (x - a) . s i n (x - b)} - \frac{c o s (x - a) s i n (x - b)}{s i n (x - a) . s i n (x - b)} d x \\ = \frac{1}{s i n (b - a)} \int [\frac{c o s (x - b)}{s i n (x - b)} - \frac{c o s (x - a)}{s i n (x - a)}] d x \\ = \frac{1}{s i n (b - a)} \int [c o t (x - b) - c o t (x - a)] d x \\ = \frac{1}{s i n (b - a)} [l o g s i n (x - b) - l o g s i n (x - a)] + C \\ = \frac{1}{s i n (b - a)} . l o g | \frac{s i n (x - b)}{s i n (x - a)} | + C \\ I = c o s e c (b - a) . l o g | \frac{s i n (x - b)}{s i n (x - a)} | + C \\ H e n c e, c o r r e c t o p t i o n i s (c) . \end{array}$

Commonly asked questions

Choose the correct option from given four options in each of the Exercises from 1 to 11.

$\int \frac{c o s 2 x - c o s 2 θ}{c o s x - c o s θ} d x$ is equal to

(A) $2 (s i n x + x c o s θ) + C$

(B) $2 (s i n x - x c o s θ) + C$

(C) $2 (s i n x + 2 x c o s θ) + C$

(D) $2 (s i n x - 2 x c o s θ) + C$

This is a Objective answer type Questions as classified in NCERT Exemplar

Sol:

$\int \frac{d x}{s i n (x - a) s i n (x - b)} i s e q u a l t o$

(A) $s i n (b - a) l o g | \frac{s i n (x - a)}{s i n (x - b)} | + C$

(B) $c o s e c (b - a) l o g | \frac{s i n (x - a)}{s i n (x - b)} | + C$

(C) $c o s e c (b - a) l o g | \frac{s i n (x - a)}{s i n (x - b)} | + C$

(D) $s i n (b - a) l o g | \frac{s i n (x - a)}{s i n (x - b)} | + C$

This is a Objective answer type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

This is a Objective answer type Questions as classified in NCERT Exemplar

$\int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} d x$

(A) $\frac{e^{x}}{1 + x^{2}}$ + C

(B) $\frac{- e^{x}}{1 + x^{2}} +$ C

(C) $\frac{e^{x}}{{(1 + x^{2})}^{2}}$ + C

(D) $\frac{- e^{x}}{{(1 + x^{2})}^{2}}$ + C

This is a Objective answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int e^{x} {(\frac{1 - x}{1 + x})}^{2} d x \\ = \int e^{x} [\frac{1 + x^{2} - 2 x}{{(1 + x^{2})}^{2}}] d x = \int e^{x} [\frac{1 + x^{2}}{{(1 + x^{2})}^{2}} - \frac{2 x}{{(1 + x^{2})}^{2}}] d x \\ = \int e^{x} [\frac{1}{1 + x^{2}} - \frac{2 x}{{(1 + x^{2})}^{2}}] d x \\ H e r e f (x) = \frac{1}{1 + x^{2}} ∴ f^{'} (x) = \frac{- 2 x}{{(1 + x^{2})}^{2}} \\ Using \int e^{x} [f (x) + f^{'} (x)] d x = e^{x} . f (x) + C \\ ∴ I = e^{x} . \frac{1}{1 + x^{2}} + C = \frac{e^{x}}{1 + x^{2}} + C \\ H e n c e, c o r r e c t o p t i o n i s (a) . \end{array}$

$\int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x$ is equal to

(A) $\frac{1}{5 x} {(4 + \frac{1}{x^{2}})}^{- 5} + C$

(B) $\frac{1}{5} {(4 + \frac{1}{x^{2}})}^{- 5} + C$

(C) $\frac{1}{10} {(1 + 4)}^{- 5} + C$

(D) $\frac{1}{10} {(\frac{1}{x^{2}} + 4)}^{- 5} + C$

This is a Objective answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x \\ = \int \frac{x^{9}}{x^{12} {(4 + \frac{1}{x^{2}})}^{6}} d x = \int \frac{1}{x^{3} {(4 + \frac{1}{x^{2}})}^{6}} d x \\ P u t (4 + \frac{1}{x^{2}}) = t \Rightarrow \frac{- 2}{x^{3}} d x = d t \Rightarrow \frac{d x}{x^{3}} = - \frac{1}{2} d t \\ ∴ I = - \frac{1}{2} \int \frac{d t}{t^{6}} = - \frac{1}{2} \times - \frac{1}{5} t^{- 5} + C = \frac{1}{10} {(4 + \frac{1}{x^{2}})}^{- 5} + C \\ H e n c e, c o r r e c t o p t i o n i s (d) . \end{array}$

If $\frac{\int d x}{(x + 2) (x^{2} + 1)} = a l o g | 1 + x^{2} | + b t a n^{- 1} (x) + \frac{1}{5} l o g | x + 2 | + C$ , then

(A) $a = \frac{- 1}{10}, b = \frac{- 2}{5}$

(B) $a = \frac{1}{10}, b = - \frac{2}{5}$

(C) $a = \frac{- 1}{10}, b = \frac{2}{5}$

(D) $a = \frac{1}{10}, b = \frac{2}{5}$

This is a Objective answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int \frac{d x}{(x + 2) (x^{2} + 1)} = a l o g | 1 + x^{2} | + b t a n^{- 1} x + \frac{1}{5} l o g | x + 2 | + C \\ L e t u s r e s o l v e t h e g i v e n integratedinto p a r t i a l f r a c t i o n s \\ P u t \frac{1}{(x + 2) (x^{2} + 1)} = \frac{A}{(x + 2)} + \frac{B x + C}{(x^{2} + 1)} \\ 1 = A (x^{2} + 1) + (x + 2) (B x + C) \\ 1 = A x^{2} + A + B x^{2} + C x + 2 B x + 2 C \\ 1 = (A + B) x^{2} + (C + 2 B) x + (A + 2 C) \\ C o m p a r i n g t h e l i k e t e r m s, w e h a v e \\ A + B = 0 \dots (i) \\ 2 B + C = 0 \dots (i i) \\ A + 2 C = 1 \dots (i i i) \\ S u b t r a c t i n g (i) f r o m (i i) w e g e t \\ 2 C - B = 1 ∴ B = 2 C - 1 \\ P u t t i n g t h e v a l u e o f B i n e q . (i i) w e h a v e \\ 2 (2 C - 1) + C = 0 \Rightarrow 4 C - 2 + C = 0 \\ 5 C = 2 ∴ C = \frac{2}{5} \\ ∴ B = 2 (\frac{2}{5}) - 1 = - \frac{1}{5} a n d A = \frac{1}{5} \\ ∴ \int \frac{d x}{(x + 2) (x^{2} + 1)} = \int \frac{\frac{1}{5}}{(x + 2)} d x + \int \frac{- \frac{1}{5} x + \frac{2}{5}}{(x^{2} + 1)} d x \\ = \frac{1}{5} \int \frac{1}{(x + 2)} d x - \frac{1}{5} \int \frac{x - 2}{(x^{2} + 1)} d x \\ = \frac{1}{5} \int \frac{1}{(x + 2)} d x - \frac{1}{5} \int \frac{x}{(x^{2} + 1)} d x + \frac{2}{5} \int \frac{1}{(x^{2} + 1)} d x \\ &thi \end{array}$

$\int \frac{x^{3}}{x + 1}$ is equal to

(A) $x + \frac{x^{2}}{2} + \frac{x^{3}}{3} - l o g | 1 - x | + C$

(B) $x + \frac{x^{2}}{2} - \frac{x^{3}}{3} - l o g | 1 + x | + C$

(C) $x - \frac{x^{2}}{2} - \frac{x^{3}}{3} - l o g | 1 + x + C$

(D) $x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - l o g | 1 - x | + C$

This is a Objective answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int \frac{x^{3}}{x + 1} d x \\ ∴ I = \int (x^{2} - x + 1 - \frac{1}{x + 1}) d x = \frac{x^{3}}{3} - \frac{x^{2}}{2} + x - l o g | x + 1 | + C \\ = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - l o g | x + 1 | + C \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

$\int \frac{x + s i n x}{1 + c o s x} d x$ is equal to

(A) $l o g | 1 + c o s x | + C$

(B) $l o g x | x + s i n x | + C$

(C) $x - t a n \frac{x}{2} + C$

(D) x $t a n \frac{x}{2} + C$

This is a Objective answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int \frac{x + s i n x}{1 + c o s x} d x \\ = \int \frac{x}{1 + c o s x} d x + \int \frac{s i n x}{1 + c o s x} d x \\ = \int \frac{x}{2 c o s^{2} \frac{x}{2}} d x + \int \frac{2 s i n \frac{x}{2} c o s \frac{x}{2}}{2 c o s^{2} \frac{x}{2}} d x \\ = \frac{1}{2} \int x . s e c^{2} \frac{x}{2} d x + \int t a n \frac{x}{2} d x \\ = \frac{1}{2} [x . \int s e c^{2} \frac{x}{2} d x - \int (D (x) . \int s e c^{2} \frac{x}{2} d x) d x] + \int t a n \frac{x}{2} d x \\ = \frac{1}{2} [x . 2 t a n \frac{x}{2} - \int 2 t a n \frac{x}{2} d x] + \int t a n \frac{x}{2} d x \\ = x t a n \frac{x}{2} - \int t a n \frac{x}{2} d x + \int t a n \frac{x}{2} d x + C \\ ∴ I = x t a n \frac{x}{2} + C \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

(A) $a = \frac{1}{3}, b = 1$

(B) $a = - \frac{1}{3}, b = 1$

(C) $a = - \frac{1}{3}, b = - 1$

(D) $a = \frac{1}{3}, b = - 1$

This is a Objective answer type Questions as classified in NCERT Exemplar

$\int^{\frac{π}{4}} \frac{d x}{1 + c o s^{} 2 x} i s e q u a l t o$

(A) 1

(B) 2

(C) 3

(D) 4

This is a Objective answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{d x}{1 + c o s 2 x} \\ = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{d x}{2 c o s^{2} x} = \frac{1}{2} \int_{- \frac{π}{4}}^{\frac{π}{4}} s e c^{2} x d x = \frac{1}{2} {[t a n x]}_{- \frac{π}{4}}^{\frac{π}{4}} \\ = \frac{1}{2} [t a n \frac{π}{4} - t a n (- \frac{π}{4})] = \frac{1}{2} [1 + 1] = \frac{1}{2} \times 2 = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Kindly consider the following

This is a Objective answer type Questions as classified in NCERT Exemplar

Integrals Fill in the Blanks

Q1. $\int^{\frac{π}{2}} c o s x e^{s i n x} d x$ is equal to _______.

Sol:

$\begin{array}{l} L e t I = \int_{0}^{\frac{π}{2}} c o s x . e^{s i n x} d x \\ P u t s i n x = t \Rightarrow c o s x d x = d t \\ W h e n x = 0 t h e n t = s i n 0 = 0; W h e n x = \frac{π}{2} t h e n t = s i n \frac{π}{2} = 1 \\ ∴ I = \int_{0}^{1} e^{t} d t = {[e^{t}]}_{0}^{1} = (e^{1} - e^{0}) = e - 1 \\ H e n c e, I = e - 1 \end{array}$

Q2. $\int \frac{x + 3}{{(x + 4)}^{2}} e^{x} d x$ =_______.

Sol:

$\begin{array}{l} L e t I = \int \frac{x + 3}{{(x + 4)}^{2}} . e^{x} d x \\ = \int \frac{x + 4 - 1}{{(x + 4)}^{2}} . e^{x} d x \\ = \int [\frac{x + 4}{{(x + 4)}^{2}} - \frac{1}{{(x + 4)}^{2}}] . e^{x} d x = \int [\frac{1}{x + 4} - \frac{1}{{(x + 4)}^{2}}] . e^{x} d x \\ P u t \frac{1}{x + 4} = t \Rightarrow - \frac{1}{{(x + 4)}^{2}} d x = d t \\ L e t f (x) = \frac{1}{x + 4} ∴ f^{'} (x) = - \frac{1}{{(x + 4)}^{2}} \\ Using \int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C \\ ∴ I = e^{x} . \frac{1}{x + 4} + C \\ H e n c e, I = \frac{e^{x}}{x + 4} + C . \end{array}$

Commonly asked questions

Kindly consider the following

This is a Fill in the Blanks type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int_{- π}^{π} s i n^{3} x . c o s^{2} x d x \\ L e t f (x) = s i n^{3} x . c o s^{2} x \\ f (- x) = s i n^{3} (- x) . c o s^{2} (- x) = - s i n^{3} x . c o s^{2} x = - f (x) \\ ∴ \int_{- π}^{π} s i n^{3} x . c o s^{2} x d x i s a n o d d f u n c t i o n . \\ ∴ I = 0 \end{array}$

Kindly consider the following

$\int^{\frac{π}{2}} c o s x e^{s i n x} d x$ is equal to _______.

This is a Fill in the Blanks type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int_{0}^{\frac{π}{2}} c o s x . e^{s i n x} d x \\ P u t s i n x = t \Rightarrow c o s x d x = d t \\ W h e n x = 0 t h e n t = s i n 0 = 0 W h e n x = \frac{π}{2} t h e n t = s i n \frac{π}{2} = 1 \\ ∴ I = \int_{0}^{1} e^{t} d t = {[e^{t}]}_{0}^{1} = (e^{1} - e^{0}) = e - 1 \\ H e n c e, I = e - 1 \end{array}$

Kindly consider the following

$\int \frac{x + 3}{{(x + 4)}^{2}} e^{x} d x$ =_______.

This is a Fill in the Blanks type Questions as classified in NCERT Exemplar

Sol:

Kindly consider the following

If: $\int^{a} \frac{1}{1 + 4 x^{2}} d x = \frac{π}{8}$ ,then, $a =$ _______.

This is a Fill in the Blanks type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int_{0}^{a} \frac{1}{1 + 4 x^{2}} d x = \frac{π}{8} \\ \Rightarrow \frac{1}{4} \int_{0}^{a} \frac{1}{(\frac{1}{4} + x^{2})} d x = \frac{π}{8} \Rightarrow \int_{0}^{a} \frac{1}{[{(\frac{1}{2})}^{2} + x^{2}]} d x = \frac{π}{2} \\ \Rightarrow \frac{1}{1 / 2} {[t a n^{- 1} \frac{x}{1 / 2}]}_{0}^{a} = \frac{π}{2} \Rightarrow 2 [t a n^{- 1} 2 a - t a n^{- 1} 0] = \frac{π}{2} \\ \Rightarrow t a n^{- 1} 2 a = \frac{π}{4} \Rightarrow 2 a = t a n \frac{π}{4} \Rightarrow 2 a = 1 \Rightarrow a = \frac{1}{2} \\ H e n c e, t h e v a l u e o f a = \frac{1}{2} . \end{array}$

Kindly consider the following

$\int \frac{s i n x}{3 + 4 c o s^{2} x} d x =_______$

This is a Fill in the Blanks type Questions as classified in NCERT Exemplar

Maths NCERT Exemplar Solutions Class 12th Chapter Seven Exam

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