Question

The value of $l={\int }_0^{\frac{\pi }{4}}\left(ta{n}^{n+1}x\right)dx+\frac{1}{2}{\int }_0^{\frac{\pi }{2}}ta{n}^{n-1}\left(x/2\right)dx$ is equal to

• A. $\frac{1}{n}$
• B. $\frac{n+2}{2n+1}$
• C. $\frac{2n-1}{n}$
• D. $\frac{2n-3}{3n-2}$

Answer 1

The correct option is A $\frac{1}{n}$

Given $I={\int }_0^{\frac{\pi }{4}}{\tan }^{n+1}xdx+\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{\tan }^{n-1}\left(\frac{x}{2}\right)dx$

Let $Z=tan\left(x\right)t=\tan \left(\frac{x}{2}\right)$

$dz={\sec }^{2}xdxdt={\sec }^2\left(\frac{x}{2}\right)\left(\frac{1}{2}\right)dx\frac{dz}{(1+z^2)}=dx\frac{dt}{(1+t^2)}=\frac{dx}{2}$

$I={\int }_0^{\frac{\pi }{4}}\left({\tan }^{n+1}x\right)dx+\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{\tan }^{n-1}\left(\frac{x}{2}\right)dx$

$I={\int }_0^1\frac{z^n+1}{1+z^2}dx+{\int }_0^1\frac{t^{n-1}}{1+t^2}dz$ Both give equal values.

$I={\int }_0^1\frac{z^{n+1}}{1+z^2}dx+{\int }_0^1\frac{z^{n-1}}{1+z^2}dzI={\int }_0^1{z}^{n-1}\frac{1+z^2}{1+z^2}dz={\int }_0^1{z}^{n-1}dzI={\left[\frac{z^n}{n}\right]}_0^1=\frac{1}{n}$

Hence the correct answer is $\frac{1}{n}$