Question

Solve: ${\int }_0^{\pi /4}{\tan }^{100}xdx+{\int }_0^{\pi /4}{\tan }^{102}xdx=....$

• A. $\frac{1}{101}$
• B. $\frac{1}{102}$
• C. $\frac{1}{100}$
• D. $101$

Answer 1

The correct option is A $\frac{1}{101}$

$I={\int }_0^{\pi /4}{\tan }^{100}xdx+{\int }_0^{\pi /4}{\tan }^{102}xdx$

$I={\int }_0^{\pi /4}{\tan }^{100}x+{\tan }^{102}xdx={\int }_0^{\pi /4}{\tan }^{100}x[1+{\tan }^{2}x]dx$

$I={\int }_0^{\pi /4}{\tan }^{100}x{\sec }^{2}xdx={\int }_0^{\pi /4}{\left[f\right(x\left)\right]}^n{f}^{\prime }\left(x\right)dx$

where $f\left(x\right)=\tan x\Rightarrow f^{\prime }\left(x\right)={\sec }^{2}x$

$n=100$

$\therefore I={\left[\frac{{\left[f\right(x\left)\right]}^{n-1}}{n+1}\right]}_0^{\pi /4}=\frac{1}{n+1}{\left\{{\left(\tan x\right)}^{n+1}\right\}}_0^{\pi /4}=\frac{1}{n+1}\{{\left(\tan \frac{\pi }{4}\right)}^{n+1}-{\left(\tan 0\right)}^{n+1}\}$

$=\frac{1}{100+1}\{{\left(1\right)}^{100+1}-0\}=\frac{1}{101}(1^{101}-0)$

$\therefore I=\frac{1}{101}$