If {I_n} = intlimits_0^{frac{pi }{4}} {{{tan | vedclass

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(B) We are given ${I_n} = \int\limits_0^{\frac{\pi }{4}} {{{	an }^n}x\,dx}$.Consider the sum ${I_n} + {I_{n - 2}} = \int\limits_0^{\frac{\pi }{4}} {(

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(B) We are given ${I_n} = \int\limits_0^{\frac{\pi }{4}} {{{	an }^n}x\,dx}$.Consider the sum ${I_n} + {I_{n - 2}} = \int\limits_0^{\frac{\pi }{4}} {(	an^n x + 	an^{n-2} x) dx}$.Factoring out ${	an^{n-2} x}$,we get ${I_n} + {I_{n - 2}} = \int\limits_0^{\frac{\pi }{4}} {	an^{n-2} x (1 + 	an^2 x) dx}$.Since $1 + 	an^2 x = \sec^2 x$,we have ${I_n} + {I_{n - 2}} = \int\limits_0^{\frac{\pi }{4}} {	an^{n-2} x \sec^2 x dx}$.Let $u = 	an x$,then $du = \sec^2 x dx$. When $x=0, u=0$ and when $x=\pi/4, u=1$.Thus,${I_n} + {I_{n - 2}} = \int\limits_0^1 {u^{n-2} du} = \left[ \frac{u^{n-1}}{n-1} ight]_0^1 = \frac{1}{n-1}$.Now,we evaluate the limit: $\mathop {\lim }\limits_{n 	o \infty } n({I_n} + {I_{n - 2}}) = \mathop {\lim }\limits_{n 	o \infty } \frac{n}{n-1}$.Dividing numerator and denominator by $n$,we get $\mathop {\lim }\limits_{n 	o \infty } \frac{1}{1 - 1/n} = 1$.

If ${I_n} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}x\,dx}$,then $\mathop {\lim }\limits_{n \to \infty } \,n({I_n} + {I_{n - 2}})$ equals

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