Question

If $\int \frac{\sin x}{\sin 4x}dx=\frac{A}{3408}[\frac{1}{\sqrt{2}}\log \left|\frac{1+\sqrt{2}u}{1-\sqrt{2}u}\right|-\frac{1}{2}\log \left|\frac{1+u}{1-u}\right|]+C$, where $u=\sin x$, then $A$ is equal to

Answer 1

$I=\int \frac{\sin x}{\sin 4x}dx=\int \frac{\sin x}{2\sin 2x\cos 2x}dx=\int \frac{\sin x}{2\left(2\sin x\cos x\right)\cos 2x}dxI=\int \frac{1}{4\cos x(1-2{\sin }^{2}x)}dx=\frac{1}{4}\int \frac{\cos xdx}{{\cos }^{2}x(1-2{\sin }^{2}x)}u=\sin xdu=\cos xdxI=\frac{1}{4}\int \frac{1}{(1-u^2)(1-2u^2)}du$

$I=\frac{1}{4}\int \frac{2(1-u^2)-(1-2u^2)}{(1-u^2)(1-2u^2)}duI=\frac{1}{4}\int [\frac{2}{(1-2u^2)}-\frac{1}{(1-u^2)}]duI=\frac{1}{2}\int \frac{1}{1-{\left(\sqrt{2}u\right)}^2}du-\frac{1}{4}\int \frac{1}{(1-u^2)}du$

We know that ;

$\int \frac{1}{(1-x^2)}dx=\frac{1}{2}\log \left|\frac{1+x}{1-x}\right|+C$

$\therefore I=\frac{1}{2}\int \frac{1}{1-{\left(\sqrt{2}u\right)}^2}du-\frac{1}{4}\int \frac{1}{(1-u^2)}duI=\frac{1}{2}\frac{1}{2\sqrt{2}}\log \left|\frac{1+\sqrt{2}u}{1-\sqrt{2}u}\right|-\frac{1}{4}\frac{1}{2}\log \left|\frac{1+u}{1-u}\right|+CI=\frac{1}{4}[\frac{1}{\sqrt{2}}\log \left|\frac{1+\sqrt{2}u}{1-\sqrt{2}u}\right|-\frac{1}{4}\log \left|\frac{1+u}{1-u}\right|]+C$

Comparing with the equation given in the question;

$\frac{A}{3408}=\frac{1}{4}\therefore A=\frac{3408}{4}=852$.