Question

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

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

Answer 1

The correct option is A $\frac{-2}{3}{\sin }^{-1}\left[{\left(\cos x\right)}^{3/2}\right]+c$

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

$=\int \sqrt{\frac{\cos x(1-{\cos }^{2}x)}{1-{\cos }^{3}x}}$ $dx$

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

Let, $\cos x=z$ then $-\sin xdx=dz$

Using this in the above integration we get,

$=-\int \sqrt{\frac{z}{1-z^3}}$ $dz$

$=-\int \frac{\sqrt{z}}{\sqrt{1-z^3}}$ $dz$

$=-\frac{2}{3}\int \frac{\frac{3}{2}\sqrt{z}}{\sqrt{1-(z^{\frac{3}{2}}{)}^2}}$ $dz$

$=-\frac{2}{3}\int \frac{d\left(z^{\frac{3}{2}}\right)}{\sqrt{1-(z^{\frac{3}{2}}{)}^2}}$

$=-\frac{2}{3}{\sin }^{-1}\left(z^{\frac{3}{2}}\right)$ $+c$

$=-\frac{2}{3}{\sin }^{-1}\left[\right(\cos x{)}^{\frac{3}{2}}]$ $+c$