Question

Find the integral $\int \frac{\sin x}{{\cos }^{2}x}dx$

Answer 1

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

Let

$t=\cos x$

$dt=-\sin xdx$

Therefore,

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

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

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

$I=\sec x+c$

Hence, this is the answer.