If {I_n} = int_0^{pi /4} {{tan ^n}theta ,dthe | vedclass

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(A) Given ${I_n} = \int_0^{\pi /4} {{	an ^n}	heta \,d	heta }$.Consider ${I_{n + 1}} + {I_{n - 1}} = \int_0^{\pi /4} {{	an ^{n + 1}}	heta \,d	het

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(A) Given ${I_n} = \int_0^{\pi /4} {{	an ^n}	heta \,d	heta }$.Consider ${I_{n + 1}} + {I_{n - 1}} = \int_0^{\pi /4} {{	an ^{n + 1}}	heta \,d	heta } + \int_0^{\pi /4} {{	an ^{n - 1}}	heta \,d	heta }$.$= \int_0^{\pi /4} {{	an ^{n - 1}}	heta (	an ^2	heta + 1)d	heta }$.$= \int_0^{\pi /4} {{	an ^{n - 1}}	heta \sec ^2	heta \,d	heta }$.Let $u = 	an 	heta$,then $du = \sec ^2	heta \,d	heta$. When $	heta = 0, u = 0$ and when $	heta = \pi /4, u = 1$.$= \int_0^1 {{u^{n - 1}}du} = \left[ \frac{u^n}{n} ight]_0^1 = \frac{1}{n}$.Therefore,$n({I_{n - 1}} + {I_{n + 1}}) = 1$.

If ${I_n} = \int_0^{\pi /4} {{\tan ^n}\theta \,d\theta ,} $ then for any positive integer $n,$ the value of $n({I_{n - 1}} + {I_{n + 1}})$ is

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