Question

Let $I_n=\int {\tan }^{n}xdx,(n>1).$ If $I_4+I_6=a{\tan }^{5}x+b{x}^5+C,$ where $C$ is a constant of integration, then the ordered pair $(a,b)$ is equal to:

• A. $(-\frac{1}{5},1)$
• B. $(\frac{1}{5},0)$
• C. $(\frac{1}{5},-1)$
• D. $(-\frac{1}{5},0)$

Answer 1

The correct option is B $(\frac{1}{5},0)$

$I_n=\int {\tan }^{n}xdx$
$=\int {\tan }^{n-2}x({\sec }^{2}x-1)dx$
$=\int {\tan }^{n-2}x{\sec }^{2}xdx-\int {\tan }^{n-2}dx$
$=\frac{{\tan }^{n-1}x}{n-1}-I_{n-2}+C$
$\Rightarrow I_n+I_{n-2}=\frac{{\tan }^{n-1}x}{n-1}+C$
$\therefore I_6+I_4=\frac{{\tan }^{5}x}{5}+C$
$\text{Given},I_4+I_6=a{\tan }^{5}x+b{x}^5+C$
$\therefore a=\frac{1}{5},b=0$