Question

${\int }^{}{e}^x\left(\frac{1-\sin x}{1-\cos x}\right)dx$ is equal to :

• A. $-e^x\tan \frac{x}{2}+c$
• B. $-e^x\cot \frac{x}{2}+c$
• C. $-\frac{1}{2}{e}^x\tan \frac{x}{2}+c$
• D. $-\frac{1}{2}{e}^x\cot \frac{x}{2}+c$

Answer 1

Answer: (B)

$-e^x\cot \frac{x}{2}+c$

Let $I=\int e^x\frac{1-\sin x}{1-\cos x}dx$

$=\int e^x\frac{1-2\sin \frac{x}{2}\cos \frac{x}{2}}{1-1+2{\sin }^2\frac{x}{2}}dx$

$=\int e^x\frac{1-2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\sin }^2\frac{x}{2}}dx$

$=\int e^x(\frac{{\csc }^2\frac{x}{2}}{2}-\cot \frac{x}{2})dx$

Let $f\left(x\right)=-\cot \frac{x}{2}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{2}{\csc }^2\frac{x}{2}$

$\Rightarrow I=\int e^x(\frac{{\csc }^2\frac{x}{2}}{2}-\cot \frac{x}{2})dx$

Using $\int e^x(f\left(x\right)+f^{\prime }\left(x\right))dx=e^{x}f\left(x\right)+C$

$\Rightarrow I=-e^x\cot \frac{x}{2}+c$