Mathematics

# Solve: $\displaystyle \int \sqrt{\dfrac{\cos x - \cos^3 x}{1 - \cos^3 x}dx} =$

##### SOLUTION
$\displaystyle\int \sqrt{\dfrac{\cos x-\cos^3x}{1-\cos^3x}}dx$
$=\displaystyle\int \sqrt{\dfrac{\cos x(1-\cos^2x)}{1-\cos^3x}}dx$
$=\displaystyle\int \sqrt{\dfrac{\cos x\sin^2x}{1-\cos^3x}}dx$
$=\displaystyle\int \dfrac{\sin x\sqrt{\cos x}}{\sqrt{1-\cos^3x}}dx$
$\cos x=t$
$-\sin xdx=dt$
$\Rightarrow -\displaystyle\int \dfrac{\sqrt{t}}{\sqrt{1-t^3}}dt$
$=-\displaystyle\int \dfrac{\sqrt{t}}{\sqrt{(1)^2-(t^{3/2})^2}}dt\times \dfrac{3}{2}\times \dfrac{2}{3}$
$\Rightarrow \dfrac{-2}{3}\displaystyle\int \dfrac{\dfrac{3}{2}\sqrt{t}}{\sqrt{(1)^2-(t^{3/2})^2}}dt$ [We know $\displaystyle\int \dfrac{2x}{\sqrt{1-x^2}}dx-\cos^{-1}x$]
$\Rightarrow \dfrac{-2}{3}\cos^{-1}(t^{3/2})+C$
$I=\dfrac{-2}{3}\cos^{-1}(\cos^{\dfrac{3}{2}}x)+C$.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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