Question

$\int \sqrt{\frac{1-\cos x}{\cos \alpha-\cos x}} d x$, 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)

$I=\int \sqrt{\frac{1-\cos x}{\cos \alpha-\cos x}} d x 0< \alpha< x <\pi$
$=\int \frac{\sqrt{2} \sin \frac{x}{2} d x}{\sqrt{2 \cos ^2 \frac{\alpha}{2}-1-2 \cos ^2 \frac{x}{2}+1}}=\int \frac{\sin \frac{x}{2} d x}{\sqrt{\cos ^2 \frac{\alpha}{2}-\cos ^2 \frac{x}{2}}}$
put $\cos \frac{x}{2}=t \Rightarrow-\frac{1}{2} \sin \frac{x}{2} d x=d t$
$\Rightarrow I=\int \frac{-2 d t}{\sqrt{\cos ^2 \frac{\alpha}{2}-t^2}}=-2 \sin ^{-1}\left(\frac{\cos \frac{x}{2}}{\cos \frac{\alpha}{2}}\right)+C$