Evaluate the integral: int frac{sin x}{sin 4x | vedclass

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Question

(B) We have $I = \int \frac{\sin x}{\sin 4x} dx = \int \frac{\sin x}{2 \sin 2x \cos 2x} dx = \int \frac{\sin x}{4 \sin x \cos x (1 - 2 \sin^2 x)} dx =

Vedclass · Accepted Answer

$\frac{1}{4\sqrt{2}} \log \left| \frac{1 + \sqrt{2} \sin x}{1 - \sqrt{2} \sin x} ight| - \frac{1}{4} \log |\sec x + 	an x| + c$

Vedclass · Answer

(B) We have $I = \int \frac{\sin x}{\sin 4x} dx = \int \frac{\sin x}{2 \sin 2x \cos 2x} dx = \int \frac{\sin x}{4 \sin x \cos x (1 - 2 \sin^2 x)} dx = \frac{1}{4} \int \frac{dx}{\cos x (1 - 2 \sin^2 x)}$.Multiply numerator and denominator by $\cos x$: $I = \frac{1}{4} \int \frac{\cos x dx}{\cos^2 x (1 - 2 \sin^2 x)} = \frac{1}{4} \int \frac{\cos x dx}{(1 - \sin^2 x)(1 - 2 \sin^2 x)}$.Let $\sin x = t$, then $\cos x dx = dt$. So $I = \frac{1}{4} \int \frac{dt}{(1 - t^2)(1 - 2t^2)}$.Using partial fractions: $\frac{1}{(1 - t^2)(1 - 2t^2)} = \frac{A}{1 - t^2} + \frac{B}{1 - 2t^2}$. Solving gives $A = -1$ and $B = 2$.$I = \frac{1}{4} \int \left( \frac{2}{1 - 2t^2} - \frac{1}{1 - t^2} ight) dt = \frac{1}{2} \int \frac{dt}{1 - (\sqrt{2}t)^2} - \frac{1}{4} \int \frac{dt}{1 - t^2}$.Using $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} ight|$ and $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} ight|$, we get:$I = \frac{1}{2} \cdot \frac{1}{2\sqrt{2}} \log \left| \frac{1 + \sqrt{2}t}{1 - \sqrt{2}t} ight| - \frac{1}{4} \cdot \frac{1}{2} \log \left| \frac{1+t}{1-t} ight| + c$.Since $\frac{1}{2} \log \left| \frac{1+\sin x}{1-\sin x} ight| = \log |\sec x + 	an x|$, the result is $\frac{1}{4\sqrt{2}} \log \left| \frac{1 + \sqrt{2} \sin x}{1 - \sqrt{2} \sin x} ight| - \frac{1}{8} \log \left| \frac{1+\sin x}{1-\sin x} ight| + c = \frac{1}{4\sqrt{2}} \log \left| \frac{1 + \sqrt{2} \sin x}{1 - \sqrt{2} \sin x} ight| - \frac{1}{4} \log |\sec x + 	an x| + c$.

Evaluate the integral: $\int \frac{\sin x}{\sin 4x} dx$

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