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

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

Answers (1)

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

Solution:

\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                    

So,

\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} \\

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

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