Question

If ${\int }_0^{\pi /4}{\tan }^{n}xdx$, then $\underset{n\to \infty }{lim}n[I_n+I_{n-2}]$ is equal to

• A. $\frac{1}{2}$
• B. $1$
• C. $0$
• D. $\infty$

Answer 1

The correct option is B $1$

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

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

Adding (1) and (2)

$I_n+I_{n-2}={\int }_0^{\frac{\pi }{4}}{\tan }^{n}xdx+{\int }_0^{\frac{\pi }{4}}{\tan }^{n-2}xdx$

$={\int }_0^{\frac{\pi }{4}}{\tan }^{n-2}xdx\times ({\sec }^{2}x-1)dx+{\int }_0^{\frac{\pi }{2}}{\tan }^{n-2}xdx$

$={\int }_0^{\frac{\pi }{4}}{\tan }^{n}x{\sec }^{2}xdx$

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

$\Rightarrow n(I_n+I_{n-2})=\frac{1}{1-\frac{1}{n}}\underset{n\to 0}{lim}\left(n\{I_n+I_{n-2}\}\right)=1$