Question

If $\int {\tan }^{6}xdx=\frac{1}{5}{\tan }^{5}x+A{\tan }^{3}x+\tan x-x+c$, then A is equal to

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $-\frac{2}{3}$
• D. $-\frac{1}{3}$

Answer 1

The correct option is D $-\frac{1}{3}$

$\int {\tan }^{6}xdx=\int {\tan }^{4}x\left({\tan }^{2}x\right)dx$
$=\int {\tan }^{4}x[{\sec }^{2}x-1]dx$
$=\int {\tan }^{4}x+\tan x-\int {\tan }^{4}xdx$
$=\frac{{\tan }^{5}x}{5}-[\int {\tan }^{2}x[{\sec }^x-1\left]dx\right]$
$=\frac{{\tan }^{5}x}{5}-[\frac{{\tan }^{3}x}{3}-\tan x+x]+c$
$=\frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}+\tan x-x+c$
So, $A=\frac{-1}{3}$