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

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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