I_n = _ ,0 ^ , /4 ^n x ,dx , then _ n - n ,[ I_n + I_ n - 2 ] equ… - Vidyadip

Question

${I_n} = \int_{\,0}^{\,\pi /4} {{{\tan }^n}x\,dx} $, then $\mathop {\lim }\limits_{n - \infty } n\,[{I_n} + {I_{n - 2}}]$ equals

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Answer

b
(b) ${I_n} = \int_0^{\pi /4} {{{\tan }^n}x} \,dx = \int_0^{\pi /4} {\,({{\sec }^2}x - 1){{\tan }^{n - 2}}x\,dx} $

$ = \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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