If {I_n} = int_{0}^{pi /4} {tan^n x} ,dx,then | vedclass

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(B) Given ${I_n} = \int_{0}^{\pi /4} {	an^n x} \,dx$.We know that $	an^2 x = \sec^2 x - 1$.So,${I_n} = \int_{0}^{\pi /4} {	an^{n-2} x (\sec^2 x - 1

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(B) Given ${I_n} = \int_{0}^{\pi /4} {	an^n x} \,dx$.We know that $	an^2 x = \sec^2 x - 1$.So,${I_n} = \int_{0}^{\pi /4} {	an^{n-2} x (\sec^2 x - 1)} \,dx$.${I_n} = \int_{0}^{\pi /4} {	an^{n-2} x \sec^2 x} \,dx - \int_{0}^{\pi /4} {	an^{n-2} x} \,dx$.${I_n} = \left[ \frac{	an^{n-1} x}{n-1} ight]_{0}^{\pi /4} - {I_{n-2}}$.Since $	an(\pi/4) = 1$ and $	an(0) = 0$,we get ${I_n} = \frac{1}{n-1} - {I_{n-2}}$.Thus,${I_n} + {I_{n-2}} = \frac{1}{n-1}$.Now,we need to find $\lim_{n 	o \infty} n[{I_n} + {I_{n-2}}]$.$\lim_{n 	o \infty} n \left( \frac{1}{n-1} ight) = \lim_{n 	o \infty} \frac{n}{n-1} = \lim_{n 	o \infty} \frac{1}{1 - 1/n} = 1$.

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

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