Question

$I_n=\int {\tan }^{n}xdx$ for $n\ge 2,$ then $I_n+I_{n-2}$ is equal to

• A. ${\tan }^{n}x+c$
• B. $\frac{({\tan }^{n-1}x)}{n-1}+c$
• C. $\frac{{\tan }^{n}x}{n}+c$
• D. $n{\tan }^{n}x+c$

Answer 1

Answer: (B)

We have, $I_n=\int ta{n}^{n}xdx$
$=\int {\tan }^{n-2}x({\tan }^{2}x)dx$
$\int {\tan }^{n-2}x({\sec }^{2}x-1)dx$
$\int {\tan }^{n-2}x.{\sec }^{2}xdx-I_{n-2}$
$\Rightarrow$ $I_n+I_{n-2}=\int {\tan }^{n-2}x.{\sec }^{2}xdx$
Put $\tan x=t$
$\Rightarrow$ ${\sec }^{2}xdx=dt$
$I_n+I_{n-2}=\int t^{n-2}dt$
$=\frac{t^{n-1}}{n-1}+c$
$\Rightarrow$ $I_n+I_{n-2}=\frac{(tan{}^{n-1}x)}{n-1}+c$