int sqrt{frac{cos x - cos^3 x}{1 - cos^3 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 $\sqrt{\sin^

Vedclass · Accepted Answer

$\frac{2}{3} \cos^{-1}(\cos^{3/2} x) + c$

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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 $\sqrt{\sin^2 x} = \sin x$ (assuming $\sin x > 0$),we have $I = \int \sqrt{\frac{\cos x}{1 - \cos^3 x}} \sin x \, dx$. Let $t = \cos x$,then $dt = -\sin x \, dx$,so $\sin x \, dx = -dt$. Substituting these into the integral: $I = - \int \sqrt{\frac{t}{1 - t^3}} \, dt = - \int \frac{\sqrt{t}}{\sqrt{1 - (t^{3/2})^2}} \, dt$. Let $u = t^{3/2}$,then $du = \frac{3}{2} t^{1/2} \, dt = \frac{3}{2} \sqrt{t} \, dt$,which implies $\sqrt{t} \, dt = \frac{2}{3} du$. Thus,$I = - \int \frac{2/3}{\sqrt{1 - u^2}} \, du = - \frac{2}{3} \sin^{-1}(u) + c = - \frac{2}{3} \sin^{-1}(t^{3/2}) + c$. Using the identity $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$,we have $-\frac{2}{3} \sin^{-1}(t^{3/2}) + c = \frac{2}{3} \cos^{-1}(t^{3/2}) + K$. Substituting $t = \cos x$,we get $I = \frac{2}{3} \cos^{-1}(\cos^{3/2} x) + c$.

$\int \sqrt{\frac{\cos x - \cos^3 x}{1 - \cos^3 x}} \, dx$ is equal to

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