Question

If $\int ta{n}^{6}xdx=\frac{1}{5}ta{n}^{5}x+Ata{n}^{3}x+tanx-x+C$ then $A$ is equal to

• A. $I/3$
• B. $2/3$
• C. $-2/3$
• D. $-1/3$

Answer 1

Answer: (D)

$-1/3$

Let $I=\int {\tan }^{6}xdx$
Using $\int {\tan }^{m}x=\frac{{\tan }^{m-1}x}{m-1}-\int {\tan }^{-2+m}dx$
$I=\frac{{\tan }^{5}x}{5}-\int {\tan }^{4}xdx$
$=\frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}+\int {\tan }^{2}xdx$
$=\frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}+\int ({\sec }^{2}x-1)dx$
$=\frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}+\tan x-x+c$

Comparing the given equation with $\int ta{n}^{6}xdx=\frac{1}{5}ta{n}^{5}x+Ata{n}^{3}x+tanx-x+C$,

we get, $A=-\frac{1}{3}$