#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 28 Maths Textbook Solution.

Answer: $\frac{\pi ^{2}}{4}-2$

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

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

Solution:

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

Integrating by parts then $x^{2}$ and cos x be the two parts

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

\begin{aligned} &=\left[\left(\frac{\pi}{2}\right)^{2} \sin \frac{\pi}{2}-0^{2} \sin 0\right]-\left\{2\left[x \int \sin x d x\right]_{0}^{\frac{\pi}{2}}-2 \int_{0}^{\frac{\pi}{2}}\left\{\frac{d(x)}{d x} \int \sin x d x\right\} d x\right\} \\ &=\left[\frac{\pi^{2}}{4}-0\right]-2[x(-\cos x)]_{0}^{\frac{\pi}{2}}+2 \int_{0}^{\frac{\pi}{2}} 1(-\cos x) d x \end{aligned}

(Again using integrating by parts method)

$\left[\begin{array}{l} \int \sin x d x=-\cos x \\ \sin \frac{\pi}{2}=1, \sin 0=0 \end{array}\right]$

\begin{aligned} &=\frac{\pi^{2}}{4}+2[x \cos x]_{0}^{\frac{\pi}{2}}-2 \int_{0}^{\frac{\pi}{2}} \cos x d x\\ &=\frac{\pi^{2}}{4}+2\left[\frac{\pi}{2} \cos \frac{\pi}{2}-0 \cos 0\right]-2[\sin x]_{0}^{\frac{\pi}{2}} \end{aligned}                                                            $\left [ \int \cos xdx=\sin x \right ]$

$=\frac{\pi^{2}}{4}+2\left[\frac{\pi}{2} 0-0 \times 1\right]-2\left[\sin \frac{\pi}{2}-\sin 0\right] \quad\left[\begin{array}{l} \cos \frac{\pi}{2}=0 \\ \cos 0=1 \end{array}\right]$

\begin{aligned} &=\frac{\pi^{2}}{4}+2 \times 0-2(1-0) \\ &=\frac{\pi^{2}}{4}-2 \end{aligned}