Question

lf $I_n={\int }_0^{\frac{\pi }{4}}$ ${\tan }^{n}xdx$, then $\underset{n\to \infty }{lim}n[I_n+I_{n+2}]=$

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

Answer 1

The correct option is B $1$

Given, $I_n={\int }_0^{\pi /4}{\tan }^{n}xdx$

and we have to find ${lim}_{n\to \infty }n[I_n+I_{n+2}]$

Let us first solve the function and get the answer in terms of $n$ and then apply limit .

Now $n[I_n+I_{n+2}]$

$=n[{\int }_0^{\pi /4}{\tan }^{n}xdx$ $+{\int }_0^{\pi /4}{\tan }^{n+2}xdx]$

$=n\left[{\int }_0^{\pi /4}\right({\tan }^{n}x+{\tan }^{n+2}x\left)dx\right]$

Since the limits of both the integrals is same it can be combined into one integral

$=n\left[{\int }_0^{\pi /4}\right({\tan }^{n}x(1+{\tan }^{2}x)dx]$

Since $1+{\tan }^{2}x={\sec }^{2}x$

Therefore, $=n\left[{\int }_0^{\pi /4}{\tan }^{n}x{\sec }^{2}x\right]$

Taking $\tan x=t$

$dx={\sec }^{2}xdt$

Upper limit becomes $\tan \left(\pi /4\right)=1$ and lower limit becomes $\tan \left(0\right)=0$

Using this the integrand becomes

$n{\int }_0^1{t}^{n}dt$

$=n(t^{n+1}/n+1{)}_0^1$

Substituting limits, we get $\frac{n}{n+1}$

Now applying limit tending to infinity to this value

${lim}_{n\to \infty }n[I_n+I_{n+2}]$

$={lim}_{n\to \infty }\frac{n}{n+1}$

$=1$