int frac{sin 2x}{sin 5x sin 3x} , dx = | vedclass

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(C) We have the integral $I = \int \frac{\sin 2x}{\sin 5x \sin 3x} \, dx$.Since $2x = 5x - 3x$,we can write $\sin 2x = \sin(5x - 3x)$.Using the identi

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$\frac{1}{3}\log |\sin 3x| - \frac{1}{5}\log |\sin 5x| + c$

Vedclass · Answer

(C) We have the integral $I = \int \frac{\sin 2x}{\sin 5x \sin 3x} \, dx$.Since $2x = 5x - 3x$,we can write $\sin 2x = \sin(5x - 3x)$.Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we get:$I = \int \frac{\sin 5x \cos 3x - \cos 5x \sin 3x}{\sin 5x \sin 3x} \, dx$$I = \int \left( \frac{\sin 5x \cos 3x}{\sin 5x \sin 3x} - \frac{\cos 5x \sin 3x}{\sin 5x \sin 3x} \right) \, dx$$I = \int \left( \cot 3x - \cot 5x \right) \, dx$Using the formula $\int \cot(ax) \, dx = \frac{1}{a} \log |\sin(ax)| + c$,we get:$I = \frac{1}{3} \log |\sin 3x| - \frac{1}{5} \log |\sin 5x| + c$.

$\int \frac{\sin 2x}{\sin 5x \sin 3x} \, dx = $

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