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

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(C) Given, $I_n = \int_0^{\pi / 4} 	an^n 	heta \, d	heta$. We need to find $I_{n-1} + I_{n+1}$. $I_{n-1} + I_{n+1} = \int_0^{\pi / 4} 	an^{n-1}

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$\frac{1}{n}$

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(C) Given, $I_n = \int_0^{\pi / 4} 	an^n 	heta \, d	heta$. We need to find $I_{n-1} + I_{n+1}$. $I_{n-1} + I_{n+1} = \int_0^{\pi / 4} 	an^{n-1} 	heta \, d	heta + \int_0^{\pi / 4} 	an^{n+1} 	heta \, d	heta$. Factor out $	an^{n-1} 	heta$: $I_{n-1} + I_{n+1} = \int_0^{\pi / 4} 	an^{n-1} 	heta (1 + 	an^2 	heta) \, d	heta$. Since $1 + 	an^2 	heta = \sec^2 	heta$, we have: $I_{n-1} + I_{n+1} = \int_0^{\pi / 4} 	an^{n-1} 	heta \sec^2 	heta \, d	heta$. Let $u = 	an 	heta$, then $du = \sec^2 	heta \, d	heta$. When $	heta = 0$, $u = 0$. When $	heta = \pi / 4$, $u = 1$. Substituting these into the integral: $I_{n-1} + I_{n+1} = \int_0^1 u^{n-1} \, du = \left[ \frac{u^n}{n} ight]_0^1 = \frac{1}{n} - 0 = \frac{1}{n}$.

If $I_n = \int_0^{\pi / 4} \tan^n \theta \, d\theta$ for $n = 1, 2, 3, \ldots$, then $I_{n-1} + I_{n+1}$ is equal to

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