Mathematics

# $\int _{ 0 }^{{ \pi /2 } }{ { x }^{ 2 } } { cos }^{ 2 }$ x dx

##### SOLUTION

We have,

$\int_{0}^{\dfrac{\pi }{2}}{{{x}^{2}}{{\cos }^{2}}}xdx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore \cos 2\theta =2{{\cos }^{2}}\theta -1$

$\Rightarrow \int_{0}^{\dfrac{\pi }{2}}{{{x}^{2}}}\dfrac{\left( 1+\cos 2x \right)}{2}dx$

$\Rightarrow \int_{0}^{\dfrac{\pi }{2}}{\dfrac{{{x}^{2}}+{{x}^{2}}\cos 2x}{2}dx}$

$\Rightarrow \dfrac{1}{2}\int_{0}^{\dfrac{\pi }{2}}{{{x}^{2}}+\dfrac{1}{2}\int_{0}^{\dfrac{\pi }{2}}{{{x}^{2}}\cos 2x}dx}$

On integration and we get,

$\dfrac{1}{2}{{\left[ \dfrac{{{x}^{3}}}{3} \right]}_{0}}^{\dfrac{\pi }{2}}+\dfrac{1}{2}\left[ {{x}^{2}}\int_{0}^{\dfrac{\pi }{2}}{\cos 2xdx}-\int_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{d}{dx}{{x}^{2}}\int_{0}^{\dfrac{\pi }{2}}{\cos 2x}dx \right)dx} \right]$

$\Rightarrow \dfrac{1}{2}\left[ \dfrac{\dfrac{{{\pi }^{3}}}{8}-{{0}^{3}}}{3} \right]+\dfrac{1}{2}\left[ {{\left\{ {{x}^{2}}\left( \dfrac{\sin 2x}{2} \right) \right\}}_{0}}^{\dfrac{\pi }{2}}-\int_{0}^{\dfrac{\pi }{2}}{\left( 2x\left( \dfrac{\sin 2x}{2} \right) \right)dx} \right]$

$\Rightarrow \dfrac{{{\pi }^{3}}}{48}+\dfrac{1}{2}\left[ \left( \dfrac{\pi }{2}-0 \right)\dfrac{\left( \sin 2\times \dfrac{\pi }{2}-\sin 0 \right)}{2} \right]-\dfrac{1}{2}\int_{0}^{\dfrac{\pi }{2}}{\left( x\sin 2x \right)dx}$

$\Rightarrow \dfrac{{{\pi }^{3}}}{48}+\dfrac{1}{2}\left[ \left( \dfrac{\pi }{2} \right)\dfrac{\left( 0-0 \right)}{2} \right]-\dfrac{1}{2}\left[ x\int_{0}^{\dfrac{\pi }{2}}{\sin 2xdx}-\int_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{d}{dx}x\int_{0}^{\dfrac{\pi }{2}}{\sin 2xdx} \right)dx} \right]$

$\Rightarrow \dfrac{{{\pi }^{3}}}{48}+0-\dfrac{1}{2}{{\left[ x\left( \dfrac{-\cos 2x}{2} \right) \right]}_{0}}^{\dfrac{\pi }{2}}+\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\left( -\cos 2x \right)}{2}dx}$

$\Rightarrow \dfrac{{{\pi }^{3}}}{48}-\dfrac{1}{2}\left[ \left\{ \left( \dfrac{\pi }{2}-0 \right)\left( \dfrac{-\cos 2\times \dfrac{\pi }{2}+\cos 0}{2} \right) \right\} \right]-\dfrac{1}{2}{{\left[ \dfrac{\sin 2x}{2} \right]}_{0}}^{\dfrac{\pi }{2}}$

$\Rightarrow \dfrac{{{\pi }^{3}}}{48}-\dfrac{1}{2}\left[ \left\{ \left( \dfrac{\pi }{2} \right)\left( \dfrac{1+1}{2} \right) \right\} \right]-\dfrac{1}{2}\left[ \dfrac{\sin 2\times \dfrac{\pi }{2}-\sin 0}{2} \right]$

$\Rightarrow \dfrac{{{\pi }^{3}}}{48}-\dfrac{\pi }{4}-\dfrac{1}{2}\left[ \dfrac{0-0}{2} \right]$

$\Rightarrow \dfrac{{{\pi }^{3}}}{48}-\dfrac{\pi }{4}$

$\Rightarrow \dfrac{\pi }{4}\left( \dfrac{{{\pi }^{2}}}{12}-1 \right)$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
Evaluate using limit of sum:
$\displaystyle \int_{1}^{3} {(x+1)^2}dx$
• A. $26$
• B. $30$
• C. $32$
• D. $28$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Fill in the blanks in the following:
$\displaystyle \int_{-a}^{a} f(x) dx = 0$ if $f$ is an ................... function.

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int _{ 0 }^{ \pi /2 }{ \log { \left( \tan { x } +\cot { x } \right) } dx }$ is equal to
• A. $\displaystyle \frac { \pi }{ 2 } \log { 2 }$
• B. $\displaystyle -\frac { \pi }{ 2 } \log { 2 }$
• C. None of these
• D. $\displaystyle \pi \log { 2 }$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
Evaluate the integral
$\displaystyle \int_{-\pi}^{\pi}\left ( x^{3}+x\cos x+tan^{5}x+2 \right )dx$
• A. $0$
• B. $\pi$
• C. $2\pi$
• D. $4\pi$

Let $\displaystyle f\left ( x \right )=\frac{\sin 2x \cdot \sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }$