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Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 5 maths textbook solution

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Hint: You must know the integration rules of trigonometric function with its limits

Given:\int_{-\pi}^{\frac{\pi}{2}} \sin ^{3} x\; d x

Solution:  \int_{-\pi}^{\frac{\pi}{2}} \sin ^{3} x\; d x

\begin{aligned} &=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x \sin x \; d x \\\\ &=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(1-\cos ^{2} x\right) \sin x\; d x \end{aligned}

=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin x\; d x+\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos ^{2} x \sin x\; d x

We know that \int \sin x \; d x=-\cos x \text { and } \int \cos x\; d x=\sin x

\begin{aligned} &=[-\cos x]_{\frac{-\pi}{2}}^{\frac{\pi}{2}}+\left[\frac{-\cos ^{3} x}{3}\right]_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \\\\ &=-\cos \frac{\pi}{2}-\left(-\cos \frac{-\pi}{2}\right)+\left[\frac{\cos ^{3}\left(\frac{\pi}{2}\right)}{3}-\frac{\cos ^{3}\left(\frac{-\pi}{2}\right)}{3}\right] \\\\ &=0+0+0-0 \\\\ &=0 \end{aligned}


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