Maths Integrals

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New answer posted•a year ago

Maths Integrals Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following

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

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Maths Integrals Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following

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

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Maths Integrals Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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

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New answer posted•a year ago

Maths Integrals Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

Upvote

New answer posted•a year ago

Maths Integrals Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Maths Integrals Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following

1 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Upvote

New answer posted•a year ago

Maths Integrals Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following

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

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Upvote

New answer posted•a year ago

Maths Integrals Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following

$\int e^{- 3 x} c o s^{3} x d x$

1 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} L e t ? ? ? ? I = ? \underset{I I}{e^{? 3 x}} \underset{I}{c o s^{3} x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = c o s^{3} x . ? e^{? 3 x} ? d x ? ? (D (c o s^{3} x) . ? e^{? 3 x} ? d x) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = c o s^{3} x . \frac{e^{? 3 x}}{? 3} ? ? (3 c o s^{2} x (? s i n x) . \frac{e^{? 3 x}}{? 3}) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? c o s^{2} x s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? (1 ? s i n^{2} x) s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + ? \underset{I}{s i n^{3} x} . \underset{I I}{e^{? 3 x}} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + s i n^{3} x ? e^{? 3 x} d x ? ? (D (s i n^{3} x) . ? e^{? 3 x} ? d x) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + s i n^{3} x . \frac{e^{? 3 x}}{? 3} ? ? (3 s i n^{2} x . c o s x . \frac{e^{? 3 x}}{? 3}) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? s i n^{2} x c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? (1 ? c o s^{2} x) c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? I = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? [s i n x . \frac{e^{? 3 x}}{? 3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? c o s x . e^{? 3 x} d x ? ? c o s^{3} x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x + s i n x . \frac{e^{? 3 x}}{3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? c o s x . e^{? 3 x} d x ? I \\ ? ? ? ? ? ? ? ? ? 2 I = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] ? [s i n x . \frac{e^{? 3 x}}{3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] + ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} ? \frac{1}{3} ? c o s x . e^{? 3 x} d x + ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? 2 I = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} + \frac{2}{3} ? c o s x . e^{? 3 x} d x \\ N o w, ? I_{1} = \frac{2}{3} ? \underset{I}{c o s x} . \underset{I I}{e^{? 3 x}} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . ? e^{? 3 x} d x ? ? (D (c o s x) . ? e^{? 3 x} ? d x) d x] \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . \frac{e^{? 3 x}}{? 3} ? ? ? s i n x . \frac{e^{? 3 x}}{? 3} d x] \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . \frac{e^{? 3 x}}{? 3} ? \frac{1}{3} ? s i n x . e^{? 3 x} d x] \end{array}$

$\begin{array}{l} = \frac{2}{? 9} c o s x . e^{? 3 x} ? \frac{2}{9} ? s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = \frac{2}{? 9} c o s x . e^{? 3 x} ? \frac{2}{9} [s i n x . \frac{e^{? 3 x}}{? 3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{2}{27} ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{1}{9} . \frac{2}{3} ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{1}{9} . I_{1} \\ I_{1} + \frac{1}{9} I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? \frac{10 I_{1}}{9} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{1}{10} c o s x . e^{? 3 x} + \frac{1}{15} s i n x . e^{? 3 x} \\ S o, ? ? ? ? ? 2 I = ? \frac{1}{3} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} ? \frac{1}{10} c o s x . e^{? 3 x} + \frac{1}{15} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? I = ? \frac{1}{6} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{6} s i n x . e^{? 3 x} ? \frac{1}{20} c o s x . e^{? 3 x} + \frac{1}{30} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{6} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{5} s i n x . e^{? 3 x} ? \frac{1}{20} c o s x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{e^{? 3 x}}{24} [s i n 3 x ? c o s 3 x] + \frac{3 e^{? 3 x}}{40} [s i n x \end{array}$