Question

$I_n={\int }_0^{\frac{\pi }{4}}ta{n}^{n}xdx,thenli{m}_{n\to \infty }n[I_n+I_{n+2}]equals$

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

Answer 1

The correct option is B $1$

$I_n+I_{n+2}={\int }_0^{\frac{\pi }{4}}ta{n}^{n}x(1+ta{n}^{2}x)dx$
$={\int }_0^{\frac{\pi }{4}}se{c}^{2}x.ta{n}^{n}xdx$
$Puttanx=t\Rightarrow se{c}^{2}xdx=dt$
$\therefore I_n+I_{n+2}={\int }_0^1{t}^{n}dt=\left[\frac{t^{n+1}}{n+1}\right]=\frac{1}{n+1}$
$\Rightarrow {lim}_{n\to \infty }n[I_n+I_{n+2}]={lim}_{n\to \infty }\frac{n}{n+1}=1$