Q94 / 225 · JEE Main 2002 · Mathematics

$I_n=\int_0^{\pi / 4} \tan ^n x d x$ then $\operatorname{Lim}_{n \rightarrow \infty} n\left[I_n+I_{n-2}\right]$ equals

  • A
    1/2
  • Correct answer
    B
    1
  • C
    $\infty$
  • D
    zero

Solution

$\int_0^{\pi / 4} \tan ^n x\left(1+\tan ^2 x\right) d x=\int_0^{\pi / 4} \tan ^n x \sec ^2 x d x=\int_0^1 t^n d t$ where $t=\tan x$ $I_n+I_{n+2}=\frac{1}{n+1} ; \Rightarrow \operatorname{Lim}_{x \rightarrow \infty} n\left[l_n+I_{n+2}\right]=\operatorname{Lim}_{x \rightarrow \infty} n \cdot \frac{1}{n+1}=\frac{n}{n+1}=\frac{n}{n\left(1+\frac{1}{n}\right)}=1$

Hence, the correct answer is option: 'B'.

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