int frac{dx}{sin x + sin 2x} = | vedclass

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Question

(A) Let $I = \int \frac{dx}{\sin x + \sin 2x} = \int \frac{dx}{\sin x(1 + 2\cos x)}$.Multiply numerator and denominator by $\sin x$:$I = \int \frac{\s

Vedclass · Accepted Answer

$\frac{1}{6}\log |1 - \cos x| + \frac{1}{2}\log |1 + \cos x| - \frac{2}{3}\log |1 + 2\cos x| + C$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{\sin x + \sin 2x} = \int \frac{dx}{\sin x(1 + 2\cos x)}$.Multiply numerator and denominator by $\sin x$:$I = \int \frac{\sin x \, dx}{\sin^2 x(1 + 2\cos x)} = \int \frac{\sin x \, dx}{(1 - \cos^2 x)(1 + 2\cos x)} = \int \frac{\sin x \, dx}{(1 - \cos x)(1 + \cos x)(1 + 2\cos x)}$.Let $\cos x = t$,then $-\sin x \, dx = dt$,so $\sin x \, dx = -dt$.$I = - \int \frac{dt}{(1 - t)(1 + t)(1 + 2t)}$.Using partial fractions:$\frac{1}{(1 - t)(1 + t)(1 + 2t)} = \frac{A}{1 - t} + \frac{B}{1 + t} + \frac{C}{1 + 2t}$.Solving for coefficients,we get $A = \frac{1}{6}$,$B = -\frac{1}{2}$,$C = \frac{4}{3}$.$I = - \int \left( \frac{1/6}{1 - t} - \frac{1/2}{1 + t} + \frac{4/3}{1 + 2t} \right) dt$.$I = - \left[ \frac{1}{6} \frac{\log |1 - t|}{-1} - \frac{1}{2} \log |1 + t| + \frac{4}{3} \cdot \frac{1}{2} \log |1 + 2t| \right] + C$.$I = \frac{1}{6} \log |1 - t| + \frac{1}{2} \log |1 + t| - \frac{2}{3} \log |1 + 2t| + C$.Substituting $t = \cos x$ back:$I = \frac{1}{6} \log |1 - \cos x| + \frac{1}{2} \log |1 + \cos x| - \frac{2}{3} \log |1 + 2\cos x| + C$.

$\int \frac{dx}{\sin x + \sin 2x} = $

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