Integrate using integration by parts $\int e^{-\frac{x}{2}}\frac{\sqrt{1-\sin{x}}}{1+\cos{x}}dx$
My Attempt
I tried taking $u=e^{-\frac{x}{2}}$ to evaluate using the general formula $\int udv=uv-\int vdu$, but I ended up with a complex trigonometric expression which I couldn't simplify further ( I can show this complex result of mine if needed)
I'm wondering if there's a suitable substitution which would simplify the integration by part process. It would be great if anyone can give me a Hint to work this integral. Thank you in advance!
Best Answer
The answer is not-that correct: we assume $-\pi/2\le x\le\pi/2$. Look at this, we first consider the half-angle transform
$$\frac{\sqrt{1-\sin{x}}}{1+\cos{x}}=\frac{\cos \frac x2-\sin \frac x2}{2\cos^2 \frac x2}$$
Notice that $$\frac{{\rm d}}{{\rm d}x}\frac{1}{\cos \frac x2}=\frac{\sin \frac x2}{2\cos^2 \frac x2}$$
So we have
\begin{align} &\int e^{-\frac{x}{2}}\frac{\sqrt{1-\sin{x}}}{1+\cos{x}}{\rm d}x=\int e^{-\frac{x}{2}}\frac{\cos \frac x2-\sin \frac x2}{2\cos^2 \frac x2}{\rm d}x\\ =&\int e^{-\frac{x}{2}}\frac{\cos \frac x2}{2\cos^2 \frac x2}{\rm d}x-\int e^{-\frac{x}{2}}\frac{\sin \frac x2}{2\cos^2 \frac x2}{\rm d}x\\ =&\int e^{-\frac{x}{2}}\frac{1}{2\cos \frac x2}{\rm d}x-e^{-\frac{x}{2}}\frac{1}{\cos\frac x2}-\int \frac 12 e^{-\frac{x}2}\frac{1}{\cos\frac x2}{\rm d}x=-e^{-\frac{x}{2}}\frac{1}{\cos\frac x2} \end{align}