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Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 16

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Answer: \frac{3}{5\sqrt{2}}

Given:  \int_{0}^{\frac{\pi}{4}} \sin 2 x \sin 3 x d x

Hint: You must know the identity 2 sin x sin y

Solution:  \int_{0}^{\frac{\pi}{4}} \sin 2 x \sin 3 x d x

\begin{aligned} &{[\therefore 2 \sin x \sin y=\cos (x-y)-\cos (x+y)]} \\ & \end{aligned}

=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \cos x-\cos 5 x d x \\

(\therefore \cos (-x)=\cos x)

\begin{aligned} &=\frac{1}{2}\left(\sin x-\frac{\sin 5 x}{5}\right)_{0}^{\frac{\pi}{4}}=\frac{1}{2}\left(\sin \frac{\pi}{4}+\frac{\sin \frac{\pi}{4}}{5}\right) \\ & \end{aligned}

=\frac{1}{2}\left(\frac{1}{\sqrt{2}}+\frac{1}{5 \sqrt{2}}\right)=\frac{1}{2} \times\left(\frac{5+1}{5 \sqrt{2}}\right) \\

=\frac{1}{2} \times \frac{6}{5 \sqrt{2}}=\frac{3}{5 \sqrt{2}}



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