If u(n) = int_0^{frac{pi}{2}} (1 + sin t)^n s | vedclass

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(C) Given $u(n) = \int_0^{\pi/2} (1 + \sin t)^n \sin 2t \, dt$. Using the identity $\sin 2t = 2 \sin t \cos t$,we have: $u(n) = 2 \int_0^{\pi/2} (1 +

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$\frac{129}{15}$

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(C) Given $u(n) = \int_0^{\pi/2} (1 + \sin t)^n \sin 2t \, dt$. Using the identity $\sin 2t = 2 \sin t \cos t$,we have: $u(n) = 2 \int_0^{\pi/2} (1 + \sin t)^n \sin t \cos t \, dt$. Let $x = 1 + \sin t$,then $dx = \cos t \, dt$. When $t = 0$,$x = 1$. When $t = \pi/2$,$x = 2$. Substituting these into the integral: $u(n) = 2 \int_1^2 x^n (x - 1) \, dx = 2 \int_1^2 (x^{n+1} - x^n) \, dx$. Integrating,we get: $u(n) = 2 \left[ \frac{x^{n+2}}{n+2} - \frac{x^{n+1}}{n+1} \right]_1^2$. For $n = 4$: $u(4) = 2 \left[ \frac{x^6}{6} - \frac{x^5}{5} \right]_1^2 = 2 \left( (\frac{2^6}{6} - \frac{2^5}{5}) - (\frac{1}{6} - \frac{1}{5}) \right)$. $u(4) = 2 \left( (\frac{64}{6} - \frac{32}{5}) - (\frac{5 - 6}{30}) \right) = 2 \left( \frac{320 - 192}{30} + \frac{1}{30} \right)$. $u(4) = 2 \left( \frac{128}{30} + \frac{1}{30} \right) = 2 \left( \frac{129}{30} \right) = \frac{129}{15}$.

If $u(n) = \int_0^{\frac{\pi}{2}} (1 + \sin t)^n \sin 2t \, dt$,where $n \in N$,then $u(4) = $

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