Question

The integral $I=\displaystyle \int sec^{3} x tan^{3} ⁡ x d x$ is equal to (where, $C$ is the constant of integration)

• A. $sec^{5} x - sec^{3} ⁡ x + C$
• B. $\frac{sec^{5} x}{5}-sec^{3} ⁡ x + C$
• C. $\frac{sec^{5} x}{5}-\frac{sec^{3} ⁡ x}{3}+C$
• D. $\frac{sec^{5} x}{5}-tan^{- 1} ⁡ x+C$

Answer 1

Answer: (C)

Let $sec x = t\Rightarrow sec ⁡ xtan ⁡ xdx=dt$
$\therefore I=\displaystyle \int \left(sec\right)^{2} x\left(tan\right)^{2}⁡x\left(sec x tan x d x\right)$
$=\displaystyle \int t^{2}\left(t^{2} - 1\right)dt$
$=\displaystyle \int \left(t^{4} - t^{2}\right)dt$
$=\frac{t^{5}}{5}-\frac{t^{3}}{3}+C$
$I=\frac{sec^{5} x}{5}-\frac{sec^{3} ⁡ x}{3}+C$