Question

Solve $\int \frac{\sin x}{1-4{\cos }^{2}x}.dx.$

Answer 1

$\int \frac{\sin x}{1-4{\cos }^{2}x}.dx.$

Put, $2\cos x=t$

$⟹-2\sin xdx=dt$

Now, $\int \frac{\sin x}{1-4{\cos }^{2}x}dx=\frac{1}{2}\int \frac{-1}{1-t^2}dt$

$=\frac{1}{2}\int \frac{1}{t^2-1}dt$

$=\frac{1}{2}\times \frac{1}{2}\log \left|\frac{t-1}{t+1}\right|+c$

$=\frac{1}{4}\log \left|\frac{2\cos x-1}{2\cos x+1}\right|+c$ (Ans)