Question

If $I_n={\int }_0^{\frac{\pi }{4}}{\tan }^{n}x\mathrm{dx}$, then the value of $I_9+I_{11}$ is

• A. $\frac{1}{9}$
• B. $\frac{1}{10}$
• C. $\frac{1}{11}$
• D. $\frac{1}{8}$

Answer 1

The correct option is B $\frac{1}{10}$

We have,
$I_n={\int }_0^{\frac{\pi }{4}}{\tan }^{n}x\mathrm{dx}\Rightarrow I_{n+2}={\int }_0^{\frac{\pi }{4}}{\tan }^{n+2}x\mathrm{dx}\therefore I_{n+2}+I_n={\int }_0^{\frac{\pi }{4}}{\tan }^{n+2}x\mathrm{dx}+{\int }_0^{\frac{\pi }{4}}{\tan }^{n}x\mathrm{dx}\Rightarrow I_{n+2}+I_n={\int }_0^{\frac{\pi }{4}}\left({\tan }^{n+2}x+{\tan }^{n}x\right)\mathrm{dx}\Rightarrow I_{n+2}+I_n={\int }_0^{\frac{\pi }{4}}{\tan }^{n}x\left({\tan }^{2}x+1\right)\mathrm{dx}\Rightarrow I_{n+2}+I_n={\int }_0^{\frac{\pi }{4}}{\tan }^{n}x\cdot {\sec }^{2}x\mathrm{dx}\mathrm{Let}\tan x=t\Rightarrow {\sec }^{2}x\mathrm{dx}=\mathrm{dt}\mathrm{Also},x=0\Rightarrow t=0\mathrm{and}x=\frac{\pi }{4}\Rightarrow t=1\Rightarrow I_{n+2}+I_n={\int }_0^1{t}^n\mathrm{dt}={\left(\frac{t^{n+1}}{n+1}\right]}_0^1=\frac{1}{n+1}\therefore I_9+I_{11}=\frac{1}{10}$