Question

If $I_n={\int }_0^{\pi /4}{\tan }^{n}xdx$ then $\underset{n\to \infty }{lim}n(I_n+I_{n-2})=$

• A. $1$
• B. $1/2$
• C. $\infty$
• D. $0$

Answer 1

The correct option is A $1$

Given : $I_n={\int }_0^{\frac{\pi }{4}}{\tan }^{n}xdx.........\left(1\right)$

$I_{n-2}={\int }_0^{\frac{\pi }{4}}{\tan }^{n-2}xdx........\left(2\right)$

Adding $\left(1\right)and\left(2\right)$

$I_n+I_{n-2}={\int }_0^{\frac{\pi }{4}}({\tan }^{n-2}x+{\tan }^{n}x)dx{I}_n+I_{n-2}={\int }_0^{\frac{\pi }{4}}\left({\tan }^{n-2}(1+{\tan }^{2}x)\right)dx{I}_n+I_{n-2}={\int }_0^{\frac{\pi }{4}}{\tan }^{n-1}x{\sec }^{2}xdx{I}_n+I_{n-2}=\frac{{\left[{\tan }^{n+1}x\right]}_0^{\frac{\pi }{4}}}{(n+1)}=\frac{1}{(n+1)}n(I_n+I_{n-2})=\frac{n}{(n+1)}\underset{n\to \infty }{lim}n(I_n+I_{n-2})=\underset{n\to \infty }{lim}\frac{n}{n(1+\frac{1}{n})}=1$