Question

$\int \sqrt{\frac{\cos x-{\cos }^{3}x}{1-{\cos }^{3}x}}dx$, $x\in (0,\frac{\pi }{2})$is equal to

• A. $\frac{2}{3}{\cos }^{-1}\left({\cos }^{\frac{3}{2}}x\right)+C$
• B. $\frac{2}{3}{\cos }^{-1}\left({\cos }^{\frac{1}{2}}x\right)+C$
• C. $\frac{3}{2}{\cos }^{-1}\left({\cos }^{\frac{3}{2}}x\right)+C$
• D. $\frac{3}{2}{\sin }^{-1}\left({\cos }^{\frac{3}{2}}x\right)+C$

Answer 1

The correct option is A $\frac{2}{3}{\cos }^{-1}\left({\cos }^{\frac{3}{2}}x\right)+C$

$\int \sqrt{\frac{\cos x-{\cos }^{3}x}{1-{\cos }^{3}x}}dx=\int \sqrt{\frac{\cos x}{1-{\cos }^{3}x}}\sin xdx$
$=-\int \sqrt{\frac{t}{1-t^3}}dt=-\int \frac{\sqrt{t}}{\sqrt{1-{\left(t^{\frac{3}{2}}\right)}^2}}dt$ where $\cos x=t,-\sin xdx=dt$
$=-\frac{2}{3}\int \frac{\frac{3}{2}\sqrt{t}}{\sqrt{1-{\left(t^{\frac{3}{2}}\right)}^2}}dt$
$=\frac{2}{3}{\cos }^{-1}\left(t^{\frac{3}{2}}\right)+C$
$=\frac{2}{3}{\cos }^{-1}\left({\cos }^{\frac{3}{2}}x\right)+C$