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

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Question

(A) Given integral is $I = \int \frac{dx}{\sin x + \sin 2x}$.Using $\sin 2x = 2 \sin x \cos x$,we have $I = \int \frac{dx}{\sin x(1 + 2 \cos x)}$.Mult

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given integral is $I = \int \frac{dx}{\sin x + \sin 2x}$.Using $\sin 2x = 2 \sin x \cos x$,we have $I = \int \frac{dx}{\sin x(1 + 2 \cos x)}$.Multiply numerator and denominator by $\sin x$: $I = \int \frac{\sin x dx}{\sin^2 x(1 + 2 \cos x)} = \int \frac{\sin x dx}{(1 - \cos^2 x)(1 + 2 \cos x)}$.Let $\cos x = t$,then $-\sin x dx = dt$. Thus,$I = -\int \frac{dt}{(1 - t^2)(1 + 2t)} = \int \frac{dt}{(t - 1)(t + 1)(2t + 1)}$.Using partial fractions: $\frac{1}{(t - 1)(t + 1)(2t + 1)} = \frac{A}{t - 1} + \frac{B}{t + 1} + \frac{C}{2t + 1}$.Solving for constants: $1 = A(t + 1)(2t + 1) + B(t - 1)(2t + 1) + C(t^2 - 1)$.For $t = 1$: $1 = A(2)(3) \Rightarrow A = \frac{1}{6}$.For $t = -1$: $1 = B(-2)(-1) \Rightarrow B = \frac{1}{2}$.For $t = -\frac{1}{2}$: $1 = C(\frac{1}{4} - 1) = C(-\frac{3}{4}) \Rightarrow C = -\frac{4}{3}$.Substituting back: $I = \int (\frac{1/6}{t - 1} + \frac{1/2}{t + 1} - \frac{4/3}{2t + 1}) dt = \frac{1}{6} \ln|t - 1| + \frac{1}{2} \ln|t + 1| - \frac{2}{3} \ln|2t + 1| + c$.Replacing $t = \cos x$: $I = \frac{1}{6} \ln|\cos x - 1| + \frac{1}{2} \ln|\cos x + 1| - \frac{2}{3} \ln|2 \cos x + 1| + c$.Since $|\cos x - 1| = |1 - \cos x|$,the result is $\frac{1}{6} \ln|1 - \cos x| + \frac{1}{2} \ln|1 + \cos x| - \frac{2}{3} \ln|1 + 2 \cos x| + c$.

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

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