Question

If $\int e^x\frac{1-\sin x}{1-\cos x}dx=f\left(x\right)+$ constant, then $f\left(x\right)$ is equal to

• A. $e^x\cot \frac{x}{2}+c$
• B. $e^{-x}\cot \frac{x}{2}+c$
• C. $-e^x\cot \frac{x}{2}+c$
• D. $-e^{-x}\cot \frac{x}{2}+c$

Answer 1

The correct option is C $-e^x\cot \frac{x}{2}+c$

$\int e^x\frac{1-\sin x}{1-\cos x}dx$
$=\int e^x\frac{1-2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\sin }^2\frac{x}{2}}dx$
$=\int \frac{1}{2}{e}^x{\csc }^2\frac{x}{2}dx-e^x\cot \frac{x}{2}dx$
$=\int \frac{1}{2}-e^x\cot \frac{x}{2}\cdot 2+e^x\cot \frac{x}{2}2dx-e^x\cot \frac{x}{2}dx+c$
$=-e^x\cot \frac{x}{2}+c$ Using($\int e^x\left(f\right(x)+f^{\prime }(x)=e^{x}f(x)$