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#### Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 36 maths textbook solution.

Answer: $\pi -2$

Hint: use indefinite integral formula and the limits to solve this integral.

Given: $\int_{0}^{\frac{\pi}{2}}x^2\sin x dx$

Solution: $\int_{0}^{\frac{\pi}{2}}x^2\sin x dx$

Applying integration by parts method, then,

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} x^{2} \sin x \mathrm{~d} x=\left[x^{2} \int \sin x \mathrm{~d} x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left\{\frac{d\left(x^{2}\right)}{d x} \int \sin x d x\right\} d x \\ &=\left[x^{2}(-\cos x)\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} 2 x(-\cos x) d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left(\frac{d\left(x^{n}\right)}{d x}=n x^{n-1}, \int \sin a x d x=\frac{-\cos a x}{a}\right) \\ &=-\left[x^{2} \cos x\right]_{0}^{\frac{\pi}{2}}+2 \int_{0}^{\frac{\pi}{2}} x \cos x d x \end{aligned}

On applying integration by parts method
\begin{aligned} &=-\left[\left(\frac{\pi}{2}\right)^{2} \cos \frac{\pi}{2}-0^{2} \cos 0\right]+2\left[\left\{x \int \cos x d x\right\}_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left\{\frac{d(x)}{d x} \int \cos x d x\right\} d x\right] \\ &=-\frac{\pi^{2}}{4} \cdot 0-0+2\left\{[x \sin x]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} 1 \cdot \sin x d x\right\} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \\ \int \cos x d x=\sin x \end{array}\right] \\ &=2\left[\frac{\pi}{2} \cdot \sin \frac{\pi}{2}-0 \cdot \sin 0\right]-2[-\cos x]_{0}^{\frac{\pi}{2}} \\ &=\pi .1+2\left[\cos \frac{\pi}{2}-\cos 0\right]=\pi+2[0-1] \\ &=\pi-2 \end{aligned} \quad\left[\begin{array}{c} \cos \frac{\pi}{2}=0 \\ \left.\sin \frac{\pi}{2}=1\right] \\ \end{array}\right.