Mathematics

# $\int _0^{\frac{\pi }{2}}\:x^2\:cos^2x\:dx$

##### SOLUTION
$\int_{0}^{\frac{\pi}{2}}{{x}^{2}{\cos}^{2}{x}dx}$
We have $\cos{2x}=2{\cos}^{2}{x}-1\Rightarrow {\cos}^{2}{x}=\dfrac{1+\cos{2x}}{2}$
$=\int_{0}^{\frac{\pi}{2}}{{x}^{2}\left(\dfrac{1+\cos{2x}}{2}\right)dx}$
$=\int_{0}^{\frac{\pi}{2}}{{x}^{2}dx}+\int_{0}^{\frac{\pi}{2}}{{x}^{2}\left(\dfrac{\cos{2x}}{2}\right)dx}$
$=\dfrac{1}{2}\left[\dfrac{{x}^{3}}{3}\right]_{0}^{\frac{\pi}{2}}+\dfrac{1}{2}\int_{0}^{\frac{\pi}{2}}{{x}^{2}\cos{2x}dx}$
$=\dfrac{1}{2}\left[\dfrac{{\pi}^{3}}{3\times 8}-0\right]+\dfrac{1}{2}{I}_{1}$
where ${I}_{1}=\int_{0}^{\frac{\pi}{2}}{{x}^{2}\cos{2x}dx}$
Take $u={x}^{2}\Rightarrow du=2xdx$ and $dv=\cos{2x}dx\Rightarrow v=\dfrac{\sin{2x}}{2}$
$\Rightarrow {I}_{1}=\int_{0}^{\frac{\pi}{2}}{{x}^{2}\cos{2x}dx}$
$=\left[\dfrac{{x}^{2}\sin{2x}}{2}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}{\dfrac{\sin{2x}}{2}2x.dx}$
$=\dfrac{1}{2}\left[{x}^{2}\sin{2x}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}{x\sin{2x}dx}$
$=\dfrac{1}{2}\left[\dfrac{{\pi}^{2}}{4}\sin{\dfrac{2\pi}{2}}\right]-\int_{0}^{\frac{\pi}{2}}{x\sin{2x}dx}$
$=0-\int_{0}^{\frac{\pi}{2}}{x\sin{2x}dx}$
$=-\int_{0}^{\frac{\pi}{2}}{x\sin{2x}dx}$
Take $u=x\Rightarrow du=dx$ and $dv=\sin{2x}dx\Rightarrow v=\dfrac{-\cos{2x}}{2}$
$=-\left\{\left[\dfrac{-x\cos{2x}}{2}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}{\dfrac{-\cos{2x}}{2}dx}\right\}$
$=-\dfrac{\dfrac{\pi}{2}\cos{\pi}}{2}+0$
$=\dfrac{\pi}{4}$
$\therefore \int_{0}^{\frac{\pi}{2}}{{x}^{2}{\cos}^{2}{x}dx}$
$=\dfrac{1}{2}\left[\dfrac{{\pi}^{3}}{3\times 8}-0\right]+\dfrac{1}{2}{I}_{1}$
$=\dfrac{{\pi}^{3}}{48}+\dfrac{1}{2}\times \dfrac{\pi}{4}$
$=\dfrac{{\pi}^{3}}{48}+\dfrac{\pi}{8}$

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

#### Realted Questions

Q1 Single Correct Hard
$\displaystyle \int_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}\sin^{2}x. \cos^{3} x dx$ is equal to
• A. $\displaystyle \dfrac{4\pi}{30}$
• B. $\displaystyle \dfrac{\pi}{30}$
• C. $\displaystyle \dfrac{1}{15}$
• D. $\displaystyle \dfrac{4}{15}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \underset{2}{\overset{e}{\int}} \, \left[\dfrac{1}{\log \, x} - \dfrac{1}{(\log \, x)^2} \right] dx$ =
• A. $e + 2 \, \log_2 e$
• B. $e - 2 \, \log_2 e$
• C. $\log_2e$
• D. $e - 2$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int \frac{2+\sqrt{x}}{\left ( x+\sqrt{x} +1\right )^{2}}dx$
• A. $\displaystyle \frac{x}{x+\sqrt{x}+1}$
• B. $\displaystyle \frac{2\sqrt{x}}{x+\sqrt{x}+1}$
• C. $\displaystyle \frac{\sqrt{2x}}{x+\sqrt{x}+1}$
• D. $\displaystyle \frac{2x}{x+\sqrt{x}+1}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Integrate $\displaystyle\int { \dfrac { { x }^{ 2 }dx }{ { x }^{ 6 }-{ a }^{ 6 } } dx }$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$