Evaluate the integral: int sqrt{frac{cos x - | vedclass

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(C) Let $I = \int \sqrt{\frac{\cos x(1 - \cos^2 x)}{1 - \cos^3 x}} \, dx = \int \sqrt{\frac{\cos x \sin^2 x}{1 - \cos^3 x}} \, dx$. Since $\sin x$ is

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$-\frac{2}{3} \sin^{-1}(\cos^{\frac{3}{2}} x)$

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(C) Let $I = \int \sqrt{\frac{\cos x(1 - \cos^2 x)}{1 - \cos^3 x}} \, dx = \int \sqrt{\frac{\cos x \sin^2 x}{1 - \cos^3 x}} \, dx$. Since $\sin x$ is under a square root,we consider the positive root $\sin x$ (assuming the domain allows). $I = \int \sin x \sqrt{\frac{\cos x}{1 - \cos^3 x}} \, dx$. Let $u = \cos^{\frac{3}{2}} x$. Then $du = \frac{3}{2} \cos^{\frac{1}{2}} x (-\sin x) \, dx$. So,$-\frac{2}{3} du = \sqrt{\cos x} \sin x \, dx$. Substituting into the integral: $I = \int \sqrt{\frac{\cos x}{1 - (\cos^{\frac{3}{2}} x)^2}} \sin x \, dx$. This simplifies to $I = \int \frac{1}{\sqrt{1 - u^2}} \left(-\frac{2}{3} du\right) = -\frac{2}{3} \sin^{-1}(u) + c$. Substituting back $u = \cos^{\frac{3}{2}} x$,we get $I = -\frac{2}{3} \sin^{-1}(\cos^{\frac{3}{2}} x) + c$.

Evaluate the integral: $\int \sqrt{\frac{\cos x - \cos^3 x}{1 - \cos^3 x}} \, dx = \text{ . . . . . . } + c$
(where,$x \in R - \{\frac{k \pi}{2} \mid k \in Z\}$)

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Evaluate the integral: $\int \sqrt{\frac{\cos x - \cos^3 x}{1 - \cos^3 x}} \, dx = \text{ . . . . . . } + c$ (where,$x \in R - \{\frac{k \pi}{2} \mid k \in Z\}$)