Question

Evaluate:
$\int \frac{\cos x}{{\sin }^{2}x-{\sin }^2\frac{\pi }{6}}dx$

Answer 1

Given,

$\begin{array}{c}I=\int \frac{\cos x}{{\sin }^{2}x-{\sin }^2\frac{\pi }{6}}dx\\ I=\int \frac{4\cos xdx}{4{\sin }^{2}x-1}\end{array}$

Let $\sin x=t$

$dt=\cos xdx$

Therefore,

$I=\int \frac{4dt}{4t^2-1}=4\int \frac{dt}{\left(2t-1\right)\left(2t+1\right)}I=\int \frac{4dt}{2}\left(\frac{1}{2t-1}-\frac{1}{2t+1}\right)=\int \frac{2dt}{2t-1}-\int \frac{2dt}{2t+1}=\ln \left(2t-1\right)-\ln \left(2t+1\right)+C=\ln \left(\frac{2t-1}{2t+1}\right)+C=\ln \left(\frac{2\sin x-1}{2\sin x+1}\right)+C$

Hence, this is the answer.