Question

Assertion :Let $I_n={\int }_0^{\frac{\pi }{4}}{\tan }^{n}xdx$ where $nϵN$.
STATEMENT-1 : ${\int }_0^{\frac{\pi }{4}}{\tan }^{4}xdx=\frac{3\pi -8}{12}$ Reason: STATEMENT-2 : $I_n+I_{n-2}=\frac{1}{n-1}$

• A. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
• B. Statement-1 is True, Statement-2 is True; Statement-2 is Not a correct explanation for Statement-1
• C. Statement-1 is True, Statement-2 is False
• D. Statement-1 is False, Statement-2 is True

Answer 1

The correct option is A Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

$I_n={\int }_0^{\frac{\pi }{4}}{\tan }^{n}xdx={\int }_0^{\frac{\pi }{4}}{\tan }^{n-2}x\left({\tan }^{2}x\right)dx$
$={\int }_0^{\frac{\pi }{4}}{\tan }^{n-2}x({\sec }^{2}x-1)dx$
$={\int }_0^{\frac{\pi }{4}}{\tan }^{n-2}x{\sec }^{2}xdx-{\int }_0^{\frac{\pi }{4}}{\tan }^{n-2}xdx$
$={\int }_0^{\frac{\pi }{4}}{\tan }^{n-2}x{\sec }^{2}xdx-I_{n-2}$
$\therefore I_n+I_{n-2}={\int }_0^{\frac{\pi }{4}}{\tan }^{n-2}x{\sec }^{2}xdx$
Substitute $t=\tan x\Rightarrow dt={\sec }^{2}xdx$
$I_n+I_{n-2}={\int }_0^1{t}^{n-2}dt={\left[\frac{t^{n-1}}{n-1}\right]}_0^1=\frac{1}{n-1}$
Now $I_0={\int }_0^{\frac{\pi }{4}}{\tan }^{0}xdx=\frac{\pi }{4}$
$I_2+I_0=\frac{1}{2-1}=\frac{1}{2}\Rightarrow I_2=\frac{1}{2}-\frac{\pi }{4}$
$I_4+I_2=\frac{1}{4-1}=\frac{1}{3}\Rightarrow I_4=\frac{1}{3}-\frac{1}{2}+\frac{\pi }{4}=\frac{3\pi -8}{12}$