Question

${\int }_0^{\pi /2}{x}^2{cos}^2$ x dx

Answer 1

We have,

${\int }_0^{\frac{\pi }{2}}{x}^2{\cos }^{2}xdx\therefore \cos 2\theta =2{\cos }^2\theta -1$

$\Rightarrow {\int }_0^{\frac{\pi }{2}}{x}^2\frac{(1+\cos 2x)}{2}dx$

$\Rightarrow {\int }_0^{\frac{\pi }{2}}\frac{x^2+x^2\cos 2x}{2}dx$

$\Rightarrow \frac{1}{2}{\int }_0^{\frac{\pi }{2}}{x}^2+\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{x}^2\cos 2xdx$

On integration and we get,

$\frac{1}{2}{{\left[\frac{x^3}{3}\right]}_0}^{\frac{\pi }{2}}+\frac{1}{2}[x^2{\int }_0^{\frac{\pi }{2}}\cos 2xdx-{\int }_0^{\frac{\pi }{2}}\left(\frac{d}{dx}{x}^2{\int }_0^{\frac{\pi }{2}}\cos 2xdx\right)dx]$

$\Rightarrow \frac{1}{2}\left[\frac{\frac{{\pi }^3}{8}-0^3}{3}\right]+\frac{1}{2}[{{\left\{x^2\left(\frac{\sin 2x}{2}\right)\right\}}_0}^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}\left(2x\left(\frac{\sin 2x}{2}\right)\right)dx]$

$\Rightarrow \frac{{\pi }^3}{48}+\frac{1}{2}\left[(\frac{\pi }{2}-0)\frac{(\sin 2\times \frac{\pi }{2}-\sin 0)}{2}\right]-\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\left(x\sin 2x\right)dx$

$\Rightarrow \frac{{\pi }^3}{48}+\frac{1}{2}\left[\left(\frac{\pi }{2}\right)\frac{(0-0)}{2}\right]-\frac{1}{2}[x{\int }_0^{\frac{\pi }{2}}\sin 2xdx-{\int }_0^{\frac{\pi }{2}}\left(\frac{d}{dx}x{\int }_0^{\frac{\pi }{2}}\sin 2xdx\right)dx]$

$\Rightarrow \frac{{\pi }^3}{48}+0-\frac{1}{2}{{\left[x\left(\frac{-\cos 2x}{2}\right)\right]}_0}^{\frac{\pi }{2}}+{\int }_0^{\frac{\pi }{2}}\frac{(-\cos 2x)}{2}dx$

$\Rightarrow \frac{{\pi }^3}{48}-\frac{1}{2}\left[\left\{(\frac{\pi }{2}-0)\left(\frac{-\cos 2\times \frac{\pi }{2}+\cos 0}{2}\right)\right\}\right]-\frac{1}{2}{{\left[\frac{\sin 2x}{2}\right]}_0}^{\frac{\pi }{2}}$

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

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

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

$\Rightarrow \frac{\pi }{4}(\frac{{\pi }^2}{12}-1)$

Hence, this is the answer.