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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!

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  • $\begingroup$ Wolfram's answer: tinyurl.com/2rvafy6h $\endgroup$ Commented Jan 29, 2022 at 4:59
  • $\begingroup$ I am shocked that this can be expressed with elementary functions. Where did this integral come from?? $\endgroup$ Commented Jan 29, 2022 at 5:19
  • $\begingroup$ @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. $\endgroup$ Commented Jan 29, 2022 at 6:32

1 Answer 1

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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}

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  • $\begingroup$ The sign in front of the last integral is +, because $d e^{-x/2}=-\frac{1}{2}e^{-x/2}$. $\endgroup$ Commented Jan 29, 2022 at 5:48
  • $\begingroup$ I got your answer but I'm differing by a negative sign. $\endgroup$ Commented Jan 29, 2022 at 5:49
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    $\begingroup$ Sorry, there is a mis-calculation, I will fix that, sorry. @AaryanPatil $\endgroup$ Commented Jan 29, 2022 at 5:53
  • $\begingroup$ Thank you for your answer. Great Help! $\endgroup$ Commented Jan 29, 2022 at 6:52

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