Question

Solve $\int e^x\frac{1-\sin x}{1-\cos x}dx$

• A. $-e^x\cot \left(\frac{x}{2}\right)$
• B. $-e^x\tan \left(\frac{x}{2}\right)$
• C. $e^x\tan \left(\frac{x}{2}\right)$
• D. $e^x\cot \left(\frac{x}{2}\right)$

Answer 1

The correct option is A $-e^x\cot \left(\frac{x}{2}\right)$

$I=\int e^x\frac{(1-\sin x)}{(1-\cos x)}dx$
$I=\int [\frac{e^x}{2{\sin }^2\left(x/2\right)}-e^x\frac{2\sin \left(x/2\right)\cos \left(x/2\right)}{2/si{n}^2\left(x/2\right)}]dx$
$=\int e^x\left(\frac{1}{2}co{\sec }^2\frac{x}{2}\right)dx-\int e^x\cot \frac{x}{2}dx.$
Integrate first by parts
$\therefore I=e^x(-\cot \frac{x}{2})-\int e^x(-\cot \frac{x}{2})dx-\int e^x\cot \frac{x}{2}dx$
$=-e^x\cot \left(\frac{x}{2}\right)$
The last two integrals cancel.