int_{frac{pi}{3}}^{frac{pi}{2}} frac{sqrt{1+c | vedclass

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Question

(C) Let $I = \int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}} d x$.We know that $\sqrt{1+\cos x} = \sqrt{2\cos^2(x/2)} = \sqr

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$\frac{3}{2}$

Vedclass · Answer

(C) Let $I = \int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}} d x$.We know that $\sqrt{1+\cos x} = \sqrt{2\cos^2(x/2)} = \sqrt{2}\cos(x/2)$ and $\sqrt{1-\cos x} = \sqrt{2\sin^2(x/2)} = \sqrt{2}\sin(x/2)$.Alternatively,multiply numerator and denominator by $\sqrt{1-\cos x}$:$I = \int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1-\cos^2 x}}{(1-\cos x)^3} d x = \int_{\pi / 3}^{\pi / 2} \frac{\sin x}{(1-\cos x)^3} d x$.Let $t = 1 - \cos x$,then $dt = \sin x \, dx$.When $x = \pi/3$,$t = 1 - \cos(\pi/3) = 1 - 1/2 = 1/2$.When $x = \pi/2$,$t = 1 - \cos(\pi/2) = 1 - 0 = 1$.Thus,$I = \int_{1/2}^{1} t^{-3} dt = \left[ \frac{t^{-2}}{-2} \right]_{1/2}^{1} = -\frac{1}{2} [1 - (1/2)^{-2}] = -\frac{1}{2} [1 - 4] = -\frac{1}{2} (-3) = \frac{3}{2}$.

$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}} d x=$

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