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#### Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 20 Maths Textbook Solution.

Answer: $\frac{2}{3}$

Hint: Use indefinite formula then put the limit to solve this integral

Given: $\int_{0}^{\frac{\pi }{2}}\sin x\sin 2xdx$

Solution:$\int_{0}^{\frac{\pi }{2}}\sin x\sin 2xdx=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\2sin x\sin 2xdx$

$\inline =\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\{\cos (x-2 x)-\cos (x+2 x)\} d x$                            $\inline \quad[2 \sin A \sin B=\cos (A-B)-\cos (A+B)]$

$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\{\cos (-x)-\cos (3 x)\} d x \quad[\cos (-\theta)=\cos \theta]$

\begin{aligned} &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\{\cos x-\cos 3 x\} d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos x d x-\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos 3 x d x \end{aligned} \quad\left[\begin{array}{l} \int \cos x d x=\sin x \\ \left.\int \cos a x d x=\frac{\sin a x}{a}\right] \end{array}\right.

\begin{aligned} &=\frac{1}{2}[\sin x]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[\frac{\sin 3 x}{3}\right]_{0}^{\frac{\pi}{2}} \\ &=\frac{1}{2}\left[\sin \frac{\pi}{2}-\sin 0\right]-\frac{1}{3 \times 2}\left[\sin \frac{3 \pi}{2}-\sin 0\right] \quad[\sin 0=0] \end{aligned}

\begin{aligned} &=\frac{1}{2} \sin \frac{\pi}{2}-\frac{1}{6} \sin \frac{3 \pi}{2} \\ &=\frac{1}{2} \sin \frac{\pi}{2}-\frac{1}{6} \sin \left(\pi+\frac{\pi}{2}\right) \\ &=\frac{1}{2} \sin \frac{\pi}{2}-\frac{1}{6} \sin \left(-\frac{\pi}{2}\right) \quad\left[\sin \frac{\pi}{2}=1\right] \end{aligned}

$=\frac{1}{2}\times 1-\frac{1}{6}\left ( -1 \right )$

$=\frac{1}{2}+\frac{1}{6}=\frac{6+2}{12}=\frac{8}{12}$

$=\frac{2}{3}$