int_{pi /3}^{pi /2} frac{sqrt{1 + cos x}}{(1 | vedclass

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Question

(B) Let $I = \int_{\pi /3}^{\pi /2} \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{5/2}} \,dx$. Multiply the numerator and denominator by $\sqrt{1 - \cos x}$:

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(B) Let $I = \int_{\pi /3}^{\pi /2} \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{5/2}} \,dx$. Multiply the numerator and denominator by $\sqrt{1 - \cos x}$: $I = \int_{\pi /3}^{\pi /2} \frac{\sqrt{1 - \cos^2 x}}{(1 - \cos x)^{5/2} \sqrt{1 - \cos x}} \,dx = \int_{\pi /3}^{\pi /2} \frac{\sin x}{(1 - \cos x)^3} \,dx$. Let $1 - \cos x = t$,then $\sin x \,dx = dt$. When $x = \frac{\pi}{3}$,$t = 1 - \cos(\frac{\pi}{3}) = 1 - \frac{1}{2} = \frac{1}{2}$. When $x = \frac{\pi}{2}$,$t = 1 - \cos(\frac{\pi}{2}) = 1 - 0 = 1$. Substituting these into the integral: $I = \int_{1/2}^{1} t^{-3} \,dt = \left[ \frac{t^{-2}}{-2} \right]_{1/2}^{1} = -\frac{1}{2} \left[ \frac{1}{t^2} \right]_{1/2}^{1}$. $I = -\frac{1}{2} \left( 1 - 4 \right) = -\frac{1}{2} (-3) = \frac{3}{2}$.

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

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