If I_n = int_0^{frac{pi}{4}} tan^n theta , dt | vedclass

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

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$\frac{1}{11}$

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(C) We are given $I_n = \int_0^{\frac{\pi}{4}} 	an^n 	heta \, d	heta$. Consider the sum $I_n + I_{n-2} = \int_0^{\frac{\pi}{4}} (	an^n 	heta + 	an^{n-2} 	heta) \, d	heta$. Factor out $	an^{n-2} 	heta$: $I_n + I_{n-2} = \int_0^{\frac{\pi}{4}} 	an^{n-2} 	heta (	an^2 	heta + 1) \, d	heta$. Since $1 + 	an^2 	heta = \sec^2 	heta$,we have: $I_n + I_{n-2} = \int_0^{\frac{\pi}{4}} 	an^{n-2} 	heta \sec^2 	heta \, d	heta$. Let $u = 	an 	heta$,then $du = \sec^2 	heta \, d	heta$. When $	heta = 0$,$u = 0$. When $	heta = \frac{\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 $n = 12$,we have $I_{12} + I_{10} = \frac{1}{12-1} = \frac{1}{11}$.

If $I_n = \int_0^{\frac{\pi}{4}} \tan^n \theta \, d\theta$,then $I_{12} + I_{10} =$

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