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

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

Answers (1)

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.


 

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