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Question Number 37188 by rahul 19 last updated on 10/Jun/18

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

Answered by ajfour last updated on 10/Jun/18

$$I=\int \sqrt{\frac{\cos x\sin {}^{2}x}{\cos {}^{3}x(\sec {}^{3}x-1)}}dx$$$$=\int \frac{\sec x\sin xdx}{\sqrt{\sec {}^{3}x-1}}$$$$=\int \frac{d\left(\sec x\right)}{\sec x\sqrt{\sec {}^{3}x-1}}$$$$let\sec x=t,then$$$$I=\int \frac{dt}{t\sqrt{t^3-1}}=\frac{1}{3}\int \frac{d\left(t^3\right)}{t^3\sqrt{t^3-1}}$$$$let{t}^3=z$$$$\Rightarrow I=\frac{1}{3}\int \frac{dz}{z\sqrt{z-1}}$$$$letz=\sec {}^2\theta \Rightarrow dz=2\sec {}^2\theta \tan \theta d\theta$$$$I=\frac{1}{3}\int \frac{2\sec {}^2\theta \tan \theta d\theta }{\sec {}^2\theta \tan \theta }$$$$=\frac{2}{3}\theta +c=\frac{2}{3}{\cos }^{-1}\left(\frac{1}{\sqrt{z}}\right)+c$$$$\mathit{I}=\frac{2}{3}{\cos }^{-1}\sqrt{\cos {}^{3}x}+c.$$$$----------------$$$$differentiatingweobtain$$$$\frac{dI}{dx}=\frac{-2}{3\sqrt{1-\cos {}^{3}x}}\times \frac{-3\cos {}^{2}x\sin x}{2\sqrt{\cos {}^{3}x}}$$$$=\sqrt{\frac{\cos x\left(\sin {}^{2}x\right)}{1-\cos {}^{3}x}}=\sqrt{\frac{\cos x-\cos {}^{3}x}{1-\cos {}^{3}x}}.$$

Answered by MJS last updated on 10/Jun/18

$$\int \sqrt{\frac{\cos x-{\cos }^{3}x}{1-{\cos }^{3}x}}dx=\int \frac{\sin x\sqrt{\cos x}}{\sqrt{1-{\cos }^{3}x}}=$$$$[t=\sqrt{{\cos }^{3}x}\to dx=-\frac{2dt}{3\sin x\sqrt{\cos x}}]$$$$=-\frac{2}{3}\int \frac{dt}{\sqrt{1-t^2}}=-\frac{2}{3}\arcsin t=$$$$=-\frac{2}{3}\arcsin \sqrt{{\cos }^{3}x}+C$$

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