\(\int\displaystyle\frac{1}{\cos^2x\,(1+\tan x)^3}\,dx =\)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{-1}{2(1+\tan x)^2}+C\), where C is constant of integration
Since \(\displaystyle\frac{1}{\cos^2 x} = \sec^2 x\) is the derivative of tan x, substitute
\(t = 1 + \tan x, \quad dt = \sec^2 x\,dx\)
Then
\( \displaystyle \int\frac{\sec^2 x}{(1+\tan x)^3}\,dx = \int t^{-3}\,dt \)
\( \displaystyle = \frac{t^{-2}}{-2} + C \)
\( \displaystyle = \frac{-1}{2(1+\tan x)^2} + C \)