Question

$\int {\tan }^{6}xdx$ is equal to
(where $C$ is integration constant)

• A. $\frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}-\tan x+x+C$
• B. $\frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}+\tan x-x+C$
• C. $\frac{{\tan }^{4}x}{4}-\frac{{\tan }^{3}x}{3}+\tan x-x+C$
• D. $\frac{{\tan }^{4}x}{4}-\frac{{\tan }^{3}x}{3}-\tan x+x+C$

Answer 1

The correct option is B $\frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}+\tan x-x+C$

$I=\int {\tan }^{6}xdx$
We know that,
$I_n=\int {\tan }^{n}xdx=\frac{{\tan }^{n-1}x}{n-1}-I_{n-2}$
$\therefore I_6=\frac{{\tan }^{5}x}{5}-I_4=\frac{{\tan }^{5}x}{5}-(\frac{{\tan }^{3}x}{3}-I_2)=\frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}+(\frac{\tan x}{1}-I_0)=\frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}+\tan x-\int dx=\frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}+\tan x-x+C$