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Answer:

$\frac{\pi ^{3}}{48}-\frac{\pi }{8}$

Hint: Use indefinite integral formula then put the limits to solve this integral

Given:

$\int_{0}^{\frac{\pi }{2}}x^{2}\cos ^{2}xdx$

Solution:

\begin{aligned} &\int_{0}^{\frac{\pi}{2}} x^{2} \cos ^{2} x d x=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 2 x^{2} \cos ^{2} x d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} x^{2}\left(2 \cos ^{2} x\right) d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} x^{2}(1+\cos 2 x) d x \end{aligned}

\begin{aligned} &\left[\because 2 \cos ^{2} \theta=1+\cos 2 \theta\right]\\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\left(x^{2}+x^{2} \cos 2 x\right) d x\\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} x^{2} d x+\frac{1}{2} \int_{0}^{\frac{\pi}{3}}\left(x^{2} \cos 2 x\right) d x \ldots \ldots . \end{aligned}    (1)

Now

\begin{aligned} &\int_{0}^{\frac{\pi}{2}} x^{2} d x=\left[\frac{x^{2+1}}{2+1}\right]_{0}^{\frac{\pi}{2}} \\ &=\left[\frac{x^{3}}{3}\right]_{0}^{\frac{\pi}{2}} \\ &=\frac{1}{3}\left[x^{3}\right]_{0}^{\frac{\pi}{2}} \\ &=\frac{1}{3}\left[\left(\frac{\pi}{2}\right)^{3}-(0)^{3}\right] \\ &=\frac{1}{3}\left[\frac{\pi^{3}}{8}-0\right] \end{aligned}

$=\frac{\pi ^{3}}{24}$                                                                        .................(ii)

And

$\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\left(x^{2} \cos 2 x\right) d x$

Integrating by parts, then

$\int_{0}^{\frac{\pi}{2}}\left(x^{2} \cos 2 x\right) d x=\left[x^{2} \int \cos 2 x d x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left\{\frac{d}{d x}\left(x^{2}\right) \int \cos 2 x d x\right\} d x$

$=\left[x^{2} \frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} 2 x^{2-1} \frac{\sin 2 x}{2} d x$

$\left[\because \int \cos a x d x=\frac{\sin a x}{a}, \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]$

\begin{aligned} &=\frac{1}{2}\left[x^{2} \sin 2 x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} x \sin 2 x d x \\ &=\frac{1}{2}\left[\left(\frac{\pi}{2}\right)^{2} \sin 2 \times \frac{\pi}{2}-(0)^{2} \sin 2 \times 0\right]-\left\{\left[x \int \sin 2 x d x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} \frac{d}{d x}(x) \int \sin 2 x d x d x\right\} \end{aligned}

[By integrating in parts method]

$=\frac{1}{2}\left[\left(\frac{\pi}{4}\right)^{2} \sin \pi-0\right]-\left[x\left(-\frac{\cos 2 x}{2}\right)\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}} 1\left(-\frac{\cos 2 x}{2}\right) x$

$\left[\because \int \sin a x d x=-\frac{\cos a x}{a}, \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]$

$=\frac{1}{2}\left[\left(\frac{\pi}{4}\right)^{2} \times 0\right]+\left[\frac{x \cos 2 x}{2}\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos 2 x d x$                                                    $\left [ \because \sin \pi =0 \right ]$

$=0+\frac{1}{2}[x \cos 2 x]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}$

$\left [ \because \int \cos axdx=\frac{\sin ax}{a} \right ]$

\begin{aligned} &=\frac{1}{2}\left[\frac{\pi}{2} \cdot \cos 2 \times \frac{\pi}{2}-0 \cdot \cos 2 \times 0\right]-\frac{1}{4}[\sin 2 x]_{0}^{\frac{\pi}{2}} \\ &=\frac{1}{2} \cdot \frac{\pi}{2} \cdot \cos \pi-\frac{1}{4}\left[\sin 2 \times \frac{\pi}{2}-\sin 2 \times 0\right] \end{aligned}

$[\because \cos \pi=-1 \& \sin \pi=0=\sin 0]$

\begin{aligned} &=\frac{\pi}{4}(-1)-\frac{1}{4}[\sin \pi-\sin 0] \\ &=-\frac{\pi}{4}-\frac{1}{4}[0-0] \\ &=-\frac{\pi}{4} \ldots \ldots . . . .(3) \end{aligned}

Putting the value of integrals from eq(2) and (3) in (1) then

\begin{aligned} &\int_{0}^{\frac{\pi}{2}}\left(x^{2} \cos 2 x\right) d x=\frac{1}{2} \times \frac{\pi}{24}^{3}+\frac{1}{2}\left(-\frac{\pi}{4}\right) \\ &=\frac{\pi^{3}}{48}-\frac{\pi}{8} \end{aligned}

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