Question

If $I_n=\underset{0}{\stackrel{x/4}{\int }}{\tan }^{n}xdx$ $(n>1$ and is an integer), then :

• A. $I_n+I_{n-2}=\frac{1}{(n+1)}$
• B. $I+I_{n-2}=\frac{1}{(n-1)}$
• C. $I_2+I_4,I_4+I_6,$... are in H.P
• D. $\frac{1}{2(n+1)}<I_n<\frac{1}{2(n-1)}$

Answer 1

Answer: (B)

$I+I_{n-2}=\frac{1}{(n-1)}$

${\tan }^{n}x={\tan }^{n-2}x{\tan }^{2}x$

$={\tan }^{n-2}x[{\sec }^{2}x-1]$

$={\tan }^{n-2}x{\sec }^{2}x-{\tan }^{n-2}x$

$\therefore I=\int {\tan }^{n}xdx=\int {\tan }^{n-2}x{\sec }^{2}xdx-\int {\tan }^{n-2}xdx$

substitute $u=\tan x\to du={\sec }^{2}xdx$

$=\int u^{n-2}du-\int {\tan }^{n-2}xdx$

$=\frac{u^{n-1}}{n-1}-\int {\tan }^{n-2}xdx$

$\int {\tan }^{n}xdx=\frac{{\tan }^{n-1}x}{n-1}-\int {\tan }^{n-2}xdx$

$\therefore I=\frac{{\tan }^{n-1}x}{n-1}-I_{n-2}$

${\int }_0^{\frac{\pi }{4}}{\tan }^{n-1}xdx=1$

$\therefore I=\frac{1}{n-1}-I_{n-2}$

$\Rightarrow I+I_{n-2}=\frac{1}{n-1}$