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 $$dv=\sin{2x}dx\Rightarrow v=\dfrac{-\cos{2x}}{2}$$
   $$=-\left\{\left[\dfrac{-x\cos{2x}}{2}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}{\dfrac{-\cos{2x}}{2}dx}\right\}$$
   $$=-\dfrac{\dfrac{\pi}{2}\cos{\pi}}{2}+0$$
   $$=\dfrac{\pi}{4}$$
$$\therefore \int_{0}^{\frac{\pi}{2}}{{x}^{2}{\cos}^{2}{x}dx}$$
   $$=\dfrac{1}{2}\left[\dfrac{{\pi}^{3}}{3\times 8}-0\right]+\dfrac{1}{2}{I}_{1}$$
    $$=\dfrac{{\pi}^{3}}{48}+\dfrac{1}{2}\times \dfrac{\pi}{4}$$
     $$=\dfrac{{\pi}^{3}}{48}+\dfrac{\pi}{8}$$


View Full Answer

Its FREE, you're just one step away


Subjective Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86
Enroll Now For FREE

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. $$\displaystyle \dfrac{4}{15}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
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. $$e - 2$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
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. $$\displaystyle \frac{2x}{x+\sqrt{x}+1}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

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

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Easy
Evaluate:
$$ \int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx} $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer