Question

$\int \sqrt{\frac{1-\cos x}{\cos \alpha -\cos x}}dx$, where $0<\alpha <x<\pi$, is equal to

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

Answer 1

Answer: (D)

$I=\int \sqrt{\frac{1-\cos x}{\cos \alpha -\cos x}}dx0<\alpha <x<\pi$
$=\int \frac{\sqrt{2}\sin \frac{x}{2}dx}{\sqrt{2{\cos }^2\frac{\alpha }{2}-1-2{\cos }^2\frac{x}{2}+1}}=\int \frac{\sin \frac{x}{2}dx}{\sqrt{{\cos }^2\frac{\alpha }{2}-{\cos }^2\frac{x}{2}}}$
put $\cos \frac{x}{2}=t\Rightarrow -\frac{1}{2}\sin \frac{x}{2}dx=dt$
$\Rightarrow I=\int \frac{-2dt}{\sqrt{{\cos }^2\frac{\alpha }{2}-t^2}}=-2{\sin }^{-1}\left(\frac{\cos \frac{x}{2}}{\cos \frac{\alpha }{2}}\right)+C$