Question

$I=\int \sec x\tan xdx$ is
equal to

• A. $\sec x+c$
• B. $\cos x+c$
• C. $\tan x+c$
• D. None of these

Answer 1

The correct option is A $\sec x+c$

$\int \sec x\tan xdx$

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

Let $\cos x=t⟹-\sin xdx=dt$

$⟹\int \frac{-1}{t^2}dt=\frac{1}{t}$

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