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Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 34 maths textbook solution

Answers (1)

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Answer:

1

Hint:

To solve this equation we suppose sin x as cos x.

Given:

\int_{0}^{\frac{\pi}{2}} x \sin x \; d x

Solution:

Let

I=\int_{0}^{\frac{\pi}{2}} x \sin x \; d x

\begin{aligned} &=\left[x \int \sin x \; d x+\int \cos x \; d x\right]_{0}^{\pi / 2} \\\\ &=[-x \cos x+\sin x]_{0}^{\pi / 2} \end{aligned}

\begin{aligned} &=-\frac{\pi}{2} \cos \frac{\pi}{2}+\sin \frac{\pi}{2}-0 \\\\ &=-\frac{\pi}{2} \cos \frac{\pi}{2}+1 \\\\ &=0+1=1 \end{aligned}

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