Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 21

Answer:  $\frac{-\pi}{4}$

Given:   $\int_{0}^{\frac{\pi}{2}} x^{2} \cos 2 x d x$

Hint: Apply integration by parts method

Solution:   $\int_{0}^{\frac{\pi}{2}} x^{2} \cos 2 x d x$

\begin{aligned} &{\left[x^{2} \frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} 2 x \frac{\sin 2 x}{2} d x} \\ & \end{aligned}

${\left[\because \int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x\right]}$

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

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

$=0-\frac{\pi}{2} \times \frac{1}{2}-0=-\frac{\pi}{4}$

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