If the integral 525 int_0^{frac{pi}{2}} sin 2 | vedclass

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(A) Let $I = \int_0^{\frac{\pi}{2}} \sin 2x \cdot (\cos x)^{\frac{11}{2}} \left(1 + (\cos x)^{\frac{5}{2}}\right)^{\frac{1}{2}} dx$. Using $\sin 2x =

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$176$

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(A) Let $I = \int_0^{\frac{\pi}{2}} \sin 2x \cdot (\cos x)^{\frac{11}{2}} \left(1 + (\cos x)^{\frac{5}{2}}\right)^{\frac{1}{2}} dx$. Using $\sin 2x = 2 \sin x \cos x$,we have $I = 2 \int_0^{\frac{\pi}{2}} \sin x \cos x \cdot (\cos x)^{\frac{11}{2}} \left(1 + (\cos x)^{\frac{5}{2}}\right)^{\frac{1}{2}} dx = 2 \int_0^{\frac{\pi}{2}} \sin x (\cos x)^{\frac{13}{2}} \left(1 + (\cos x)^{\frac{5}{2}}\right)^{\frac{1}{2}} dx$. Let $\cos x = t^2$,then $-\sin x dx = 2t dt$. When $x=0, t=1$; when $x=\frac{\pi}{2}, t=0$. $I = 2 \int_1^0 (t^2)^{\frac{13}{2}} (1 + (t^2)^{\frac{5}{2}})^{\frac{1}{2}} (-2t dt) = 4 \int_0^1 t^{13} (1 + t^5)^{\frac{1}{2}} t dt = 4 \int_0^1 t^{14} \sqrt{1+t^5} dt$. Let $1+t^5 = k^2$,then $5t^4 dt = 2k dk$. When $t=0, k=1$; when $t=1, k=\sqrt{2}$. Also $t^5 = k^2-1$,so $t^{10} = (k^2-1)^2$. $I = 4 \int_1^{\sqrt{2}} (k^2-1)^2 \cdot k \cdot \frac{2k}{5} dk = \frac{8}{5} \int_1^{\sqrt{2}} (k^6 - 2k^4 + k^2) dk$. $I = \frac{8}{5} \left[ \frac{k^7}{7} - \frac{2k^5}{5} + \frac{k^3}{3} \right]_1^{\sqrt{2}} = \frac{8}{5} \left[ (\frac{8\sqrt{2}}{7} - \frac{8\sqrt{2}}{5} + \frac{2\sqrt{2}}{3}) - (\frac{1}{7} - \frac{2}{5} + \frac{1}{3}) \right]$. $I = \frac{8}{5} \left[ \frac{120\sqrt{2} - 168\sqrt{2} + 70\sqrt{2}}{105} - \frac{15 - 42 + 35}{105} \right] = \frac{8}{5} \left[ \frac{22\sqrt{2}}{105} - \frac{8}{105} \right] = \frac{176\sqrt{2} - 64}{525}$. Thus,$525 I = 176\sqrt{2} - 64$. Comparing with $n\sqrt{2} - 64$,we get $n = 176$.

If the integral $525 \int_0^{\frac{\pi}{2}} \sin 2 x \cos^{\frac{11}{2}} x \left(1+\cos^{\frac{5}{2}} x\right)^{\frac{1}{2}} d x$ is equal to $(n \sqrt{2}-64)$,then $n$ is equal to

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