MCQ
${I_n} = \int_{\,0}^{\,\pi /4} {{{\tan }^n}x\,dx} $, then $\mathop {\lim }\limits_{n - \infty } n\,[{I_n} + {I_{n - 2}}]$ equals
- A$1/2$
- ✓$1$
- C$\infty $
- D$0$
$ = \int_0^{\pi /4} {{{\sec }^2}x{{\tan }^{n - 2}}} x\,dx - \int_0^{\pi /4} {{{\tan }^{n - 2}}} x\,dx$
$ = \left[ {\frac{{{{\tan }^{n - 1}}x}}{{n - 1}}} \right]_0^{\pi /4} - {I_{n - 2}}$
==> ${I_n} + {I_{n - 2}} = \frac{1}{{n - \,1}}$
Now, $\mathop {\lim }\limits_{n \to \infty } \,n[{I_n} + {I_{n - 2}}]$
$ = \mathop {\lim }\limits_{n \to \infty } \,\frac{n}{{n - 1}}$
$= \mathop {\lim }\limits_{n \to \infty } \,\frac{1}{{1 - \frac{1}{n}}} = 1$.
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