Question

If $I_n={\int }_0^{\pi /4}{\tan }^{n}xdx$, then for $n>1,I_{2n}+I_{2n-2}$ equals

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

Answer 1

Answer: (A)

$\frac{1}{2n-1}$

$I_n={\int }_0^{\frac{\pi }{4}}{\tan }^{n}xdx={\int }_0^{\frac{\pi }{4}}{\tan }^{n-2}x({\sec }^{n-2}x-1)dx$
$={\int }_0^{\frac{\pi }{4}}{\tan }^{n-2}x{\sec }^{2}xdx-{\int }_0^{\frac{\pi }{4}}{\tan }^{n-2}xdx$
$I_n={\int }_0^{\frac{\pi }{4}}{\tan }^{n-2}x{\sec }^{2}xdx-I_{n-2}$
$\therefore I_n+I_{n-2}={\int }_0^{\frac{\pi }{4}}{\tan }^{n-2}x{\sec }^{2}xdx$
Substitute $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$\therefore I_n+I_{n-2}={\int }_0^1{t}^{n-2}dt$
$={\left[\frac{t^{n-2}}{n-1}\right]}_0^1=\frac{1}{n-1}$
Hence, $I_{2n}+I_{2n-2}=\frac{1}{2n-1}$