Evaluate the integral: int frac{dx}{sin x + s | vedclass

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(B) Given $I = \int \frac{dx}{\sin x + \sin 2x}$.Using $\sin 2x = 2 \sin x \cos x$,we get $I = \int \frac{dx}{\sin x(1 + 2 \cos x)}$.Multiply numerato

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$\frac{1}{6} \log (1-\cos x) - \frac{1}{2} \log (1+\cos x) - \frac{2}{3} \log |1+2 \cos x| + c$

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(B) Given $I = \int \frac{dx}{\sin x + \sin 2x}$.Using $\sin 2x = 2 \sin x \cos x$,we get $I = \int \frac{dx}{\sin x(1 + 2 \cos x)}$.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 $I = -\int \frac{dt}{(1-t)(1+t)(1+2t)}$.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 constants: $A = \frac{1}{6}$,$B = \frac{1}{2}$,$C = -\frac{4}{3}$.Thus,$I = -\int (\frac{1/6}{1-t} + \frac{1/2}{1+t} - \frac{4/3}{1+2t}) dt = \frac{1}{6} \log |1-t| - \frac{1}{2} \log |1+t| + \frac{4}{3} \cdot \frac{1}{2} \log |1+2t| + c$.Substituting $t = \cos x$: $I = \frac{1}{6} \log |1-\cos x| - \frac{1}{2} \log |1+\cos x| + \frac{2}{3} \log |1+2 \cos x| + c$.

Evaluate the integral: $\int \frac{dx}{\sin x + \sin 2x}$

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