Question

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

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

Answer 1

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

$\int e^{\frac{-x}{2}}\frac{(\cos \frac{x}{2}-\sin \frac{x}{2})}{{\cos }^2\frac{x}{2}}d\left(\frac{x}{2}\right)$
$=\int e^{\frac{-x}{2}}(\sec \frac{x}{2}-\tan \frac{x}{2}\sec \frac{x}{2})d\left(\frac{x}{2}\right)$
Let $\frac{-x}{2}=t$
$-\int e^t(\sec t+\tan t\sec t)dt$
In the form of $\int e^x\left(f\right(x)+f^1(x\left)\right)dx=e^{x}f\left(x\right)+c$
$-e^t\sec t+c$
$=-e^{\frac{-x}{2}}\sec \frac{x}{2}+c$