int frac{sin 7x}{sin 2x sin 5x} dx = | vedclass

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Question

(C) We have the integral $I = \int \frac{\sin 7x}{\sin 2x \sin 5x} dx$. Using the identity $\sin 7x = \sin(5x + 2x)$,we can write: $I = \int \frac{\si

Vedclass · Accepted Answer

$\frac{1}{5} \log \sin 5x + \frac{1}{2} \log \sin 2x + c$

Vedclass · Answer

(C) We have the integral $I = \int \frac{\sin 7x}{\sin 2x \sin 5x} dx$. Using the identity $\sin 7x = \sin(5x + 2x)$,we can write: $I = \int \frac{\sin(5x + 2x)}{\sin 2x \sin 5x} dx$. Applying the formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$: $I = \int \frac{\sin 5x \cos 2x + \cos 5x \sin 2x}{\sin 2x \sin 5x} dx$. Dividing each term in the numerator by the denominator: $I = \int \frac{\sin 5x \cos 2x}{\sin 2x \sin 5x} dx + \int \frac{\cos 5x \sin 2x}{\sin 2x \sin 5x} dx$. $I = \int \cot 2x dx + \int \cot 5x dx$. Using the standard integral $\int \cot(ax) dx = \frac{1}{a} \log |\sin(ax)| + c$: $I = \frac{1}{2} \log |\sin 2x| + \frac{1}{5} \log |\sin 5x| + c$.

$\int \frac{\sin 7x}{\sin 2x \sin 5x} dx =$

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