Question

$\int \sqrt{1-\cos x}dx=\_\_\_\_\_\_\_+C;2\pi <x<3\pi$

• A. $-2\sqrt{2}\cos \frac{x}{2}$
• B. $-\sqrt{2}\cos \frac{x}{2}$
• C. $2\sqrt{2}\cos \frac{x}{2}$
• D. $\frac{-1}{2}\sqrt{2}\cos \frac{x}{2}$

Answer 1

Answer: (A)

$-2\sqrt{2}\cos \frac{x}{2}$

We have,

$I=\int \sqrt{1-\cos x}dx$

$I=\int \sqrt{\left({\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}-{\cos }^2\frac{x}{2}+{\sin }^2\frac{x}{2}\right)}dx$

$I=\int \sqrt{2{\sin }^2\frac{x}{2}}dx$

$I=\sqrt{2}\int \sin \frac{x}{2}dx$

$I=\sqrt{2}\frac{1}{\frac{1}{2}}(-\cos \frac{x}{2})+C$

$I=-2\sqrt{2}\cos \frac{x}{2}+C$

Hence, this is the answer.