Question

Solve $\int \frac{\cos 2x}{{\sin }^{2}x{\cos }^{2}x}dx$

Answer 1

Consider the given integral.

$I=\int \frac{\cos 2x}{{\sin }^{2}x{\cos }^{2}x}dx$

We know that

$\cos 2x={\cos }^{2}x-{\sin }^{2}x$

Therefore,

$I=\int \frac{{\cos }^{2}x-{\sin }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx$ ……… (1)

Let

$t=\sin x\cos x$

$\frac{dt}{dx}=\cos x\left(\cos x\right)+\sin x(-\sin x)$

$dt=({\cos }^{2}x-{\sin }^{2}x)dx$

Therefore,

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

$I=-\frac{1}{t}+C$

On putting the value of $t$, we get

$I=-\frac{1}{\sin x\cos x}+C$

Hence, this is the answer.