# Integrate using integration by parts $\int e^{-\frac{x}{2}}\frac{\sqrt{1-\sin{x}}}{1+\cos{x}}dx$

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!

• Wolfram's answer: tinyurl.com/2rvafy6h Jan 29 at 4:59
• I am shocked that this can be expressed with elementary functions. Where did this integral come from?? Jan 29 at 5:19
• @intellect4 $\frac{x}{2} = \arcsin t$ gives an exponential with an arcsin, times rational functions of square roots. Integration by parts in this domain naturally cancels out powers of square root binomials. Jan 29 at 6:32

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}

• The sign in front of the last integral is +, because $d e^{-x/2}=-\frac{1}{2}e^{-x/2}$. Jan 29 at 5:48
• I got your answer but I'm differing by a negative sign. Jan 29 at 5:49
• Sorry, there is a mis-calculation, I will fix that, sorry. @AaryanPatil Jan 29 at 5:53
• Thank you for your answer. Great Help!
– emil
Jan 29 at 6:52