Question

Evaluate : $\int {\left\{\frac{\cos x-{\cos }^{3}x}{(1-{\cos }^{3}x)}\right\}}^{1/2}dx$

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

Answer 1

The correct option is B $\frac{2}{3}{\sin }^{-1}\left({\cos }^{3/2}x\right)$

Let $I=\int {\left\{\frac{\cos x-{\cos }^{3}x}{(1-{\cos }^{3}x)}\right\}}^{1/2}dx$

$I=\int {\left\{\frac{\cos x(1-{\cos }^{2}x)}{(1-{\cos }^{3}x)}\right\}}^{1/2}dx$

$=\int \frac{\sin x.\sqrt{\cos x}}{\sqrt{\{1-{\left({\cos }^{3/2}x\right)}^2\}}}dx$
Substitute ${\cos }^{3/2}x=t\Rightarrow -\frac{3}{2}\sin x\sqrt{\cos x}dx=dt$

Therefore

$I=-\frac{2}{3}\int \frac{1}{\sqrt{1-t^2}}dt$

We know that, $\int \frac{1}{\sqrt{a^2-x^2}}dx={\sin }^{-1}\left(\frac{x}{a}\right)+c$

$I=-\frac{2}{3}{\sin }^{-1}t=-\frac{2}{3}{\sin }^{-1}\left({\cos }^{3/2}x\right)$