Question

Evaluate:

$\int \sec x\tan xdx$

Answer 1

$\int \sec x\tan xdx$

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

$t=\cos x⟹-\sin xdx=dt$

Replacing we get,

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

$=\frac{1}{t}$

$=\frac{1}{\cos x}=\sec x$