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  • Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 18

Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 18

Answers (1)

Answer:

2

Given:

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

Hint:

You must know about \left | x \right | function and \int \sin x\; dx
 

Explanation:  

Let

I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin |x| d x

We know that

\begin{aligned} &|x|=\left\{\begin{aligned} x, & x \geq 0 \\ -x, & x<0 \end{aligned}\right. \\\\ &I=\int_{-\frac{\pi}{2}}^{0} \sin (-x) d x+\int_{0}^{\frac{\pi}{2}} \sin x \; d x \end{aligned}

\begin{aligned} &I=-\int_{-\frac{\pi}{2}}^{0} \sin x \: d x+\int_{0}^{\frac{\pi}{2}} \sin x \; d x \\\\ &=-[-\cos x]_{-\frac{\pi}{2}}^{0}+[-\cos x]_{0}^{\frac{\pi}{2}} \end{aligned}

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

 

Posted by

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