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Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise 19.3 Question 22

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Answer:  \frac{\pi}{8}+\frac{1}{4}

Hint: You must know the rules of solving definite integral.

Given:  \int_{\frac{-\pi}{4}}^{\frac{\pi}{2}} \sin x|\sin x| d x


\begin{aligned} &\mathrm{I}=\int_{\frac{-\pi}{4}}^{\frac{\pi}{2}} \sin x|\sin x| d x \\ & \end{aligned}

I=-\int_{\frac{-\pi}{4}}^{0} \sin ^{2} x d x+\int_{0}^{\frac{\pi}{2}} \sin ^{2} x d x

We will use the formula                    


\begin{aligned} &x=\frac{1-\cos 2 x}{2} \\ & \end{aligned}

=-\int_{\frac{-\pi}{4}}^{0} \frac{1-\cos 2 x}{2} d x+\int_{0}^{\frac{\pi}{2}}\left(\frac{1-\cos 2 x}{2}\right) d x

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

=-\frac{1}{2}\left(0-\left(-\frac{\pi}{4}+\frac{1}{2}\right)+\frac{1}{2}\left(\frac{\pi}{2}-0\right)\right. \\

=-\frac{\pi}{8}+\frac{1}{4}+\frac{\pi}{4} \\


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