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

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(D) We are given ${I_n} = \int_0^{\pi /4} {{	an ^n}	heta \,d	heta }$. Consider ${I_n} + {I_{n-2}} = \int_0^{\pi /4} {{	an ^n}	heta \,d	heta } +

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$\frac{1}{7}$

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(D) We are given ${I_n} = \int_0^{\pi /4} {{	an ^n}	heta \,d	heta }$. Consider ${I_n} + {I_{n-2}} = \int_0^{\pi /4} {{	an ^n}	heta \,d	heta } + \int_0^{\pi /4} {{	an ^{n-2}}	heta \,d	heta }$. ${I_n} + {I_{n-2}} = \int_0^{\pi /4} {{	an ^{n-2}}	heta (	an ^2	heta + 1) \,d	heta }$. Since $1 + 	an ^2	heta = \sec ^2	heta$,we have: ${I_n} + {I_{n-2}} = \int_0^{\pi /4} {{	an ^{n-2}}	heta \sec ^2	heta \,d	heta }$. Let $u = 	an 	heta$,then $du = \sec ^2	heta \,d	heta$. When $	heta = 0, u = 0$ and when $	heta = \pi /4, u = 1$. Thus,${I_n} + {I_{n-2}} = \int_0^1 {u^{n-2}} \,du = \left[ \frac{u^{n-1}}{n-1} ight]_0^1 = \frac{1}{n-1}$. For $n = 8$,we have ${I_8} + {I_6} = \frac{1}{8-1} = \frac{1}{7}$.

If ${I_n} = \int_0^{\pi /4} {{\tan ^n}\theta \,d\theta }$,then ${I_8} + {I_6}$ equals

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