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

Question

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

Aicharya · Accepted Answer

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

$I_n=\int_0^{\pi / 4} \tan ^n x d x$ then $\operatorname{Lim}_{n \rightarrow \infty}…

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