Question

$\int \tan ^3 2 x \sec 2 x d x$ is equal to :

• A. $\frac{1}{3} \sec ^3 2 x-\frac{1}{2} \sec 2 x+c$
• B. $-\frac{1}{6} \sec ^3 2 x-\frac{1}{2} \sec 2 x+c$
• C. $\frac{1}{6} \sec ^3 2 x-\frac{1}{2} \sec 2 x+c$
• D. $\frac{1}{3} \sec ^3 2 x+\frac{1}{2} \sec 2 x+c$

Answer 1

Answer: (C)

$\int \tan ^3 2 x \cdot \sec 2 x d x$
$I=\int\left(\sec ^2 2 x-1\right) \sec 2 x \cdot \tan 2 x d x $
Put $\sec 2 x=t \Rightarrow 2 \sec 2 x \tan 2 x d x=d t$
$\Rightarrow I=\frac{1}{2} \int\left(t^2-1\right) d t$
$=\frac{t^3}{6}-\frac{t}{2}+C=\frac{1}{6} \sec ^3 2 x-\frac{1}{2} \sec 2 x+C$