Question

$\int (sec x)^m (tan^3 x + tan x) dx$ is equal to

• A. $sec^{m+2}x+C$
• B. $sec^{m-2}x+C$
• C. $\frac{sec^{m+2}x}{m+2}+C$
• D. $\frac{tan^{m+2}x}{m+2}+C$
• E. $\frac{sec^{m+1}x}{m-1}+C$

Answer 1

Answer: (C)

Let $I=\int(\sec x)^{m}\left(\tan ^{3} x+\tan x\right) d x$
$=\int \sec ^{m} x \tan x\left(\tan ^{2} x+1\right) d x$
$=\int \sec ^{m+1} x \cdot(\sec x \tan x) d x$
Put sec $x=t \Rightarrow \sec x \tan x d x=d t=\int t^{m+1} d t$
$\quad=\frac{\sec ^{m+2} x}{m+2}+C$