Question

State whether the statement is ture/false.

${\int }_{-\pi /2}^{\pi /2}\left(\frac{\sin x}{1-\cos x}\right)dx$=0

• A. True
• B. False

Answer 1

The correct option is A True

${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\sin x}{1-\cos x}dx$

$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\sin }^2\frac{x}{2}}dx$

$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}dx$

$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cot \frac{x}{2}dx$

$=2{\left[\ln \left|\sin \frac{x}{2}\right|\right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}$

$=2[\ln \left|\sin \frac{\pi }{4}\right|-\ln \left|\sin \frac{\pi }{4}\right|]=0$