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a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following

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

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

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Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following

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alok kumar singh

Contributor-Level 10

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

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following

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

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following

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

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following

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

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alok kumar singh

Contributor-Level 10

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

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following

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alok kumar singh

Contributor-Level 10

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

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following

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

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following

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

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alok kumar singh

Contributor-Level 10

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

Sol:

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