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I tried to solve the integral $\int \frac{1}{\sqrt{1+\sin2x}}dx$ by using $1+\sin2x=(\sin x+\cos x)^2$ but got stuck. So I referred the solution in my book which is given below:

$$ I=\int \frac{1}{\sqrt{1+\sin2x}}dx $$ $$ = \int \frac{1}{\sqrt{1-\cos(\frac \pi 2+2x)}}dx\tag1\\ = \int \frac{1}{\sqrt{2\sin^2(\frac \pi 4+x)}}dx\\ =\frac 1 {\sqrt2} \int \csc\left(\frac \pi 4+x\right)dx\\ = \color{red}{\frac 1 {\sqrt2} \log\left|\tan\left(\frac \pi 8+\frac x 2\right)\right|+C}(*) $$

I wondered what if I replaced $\sin 2x$ by $\cos\left(\frac \pi 2 -2x\right)$ instead of $-\cos\left(\frac \pi 2 +2x\right)$ in step $(1)$. So I proceeded as follows:

$$ I=\int \frac{1}{\sqrt{1+\sin2x}}dx $$ $$ = \int \frac{1}{\sqrt{1+\cos(\frac \pi 2-2x)}}dx\\ = \int \frac{1}{\sqrt{2\cos^2(\frac \pi 4-x)}}dx\\ =\frac 1 {\sqrt2} \int \sec\left(\frac \pi 4-x\right)dx\\ = \color{red}{\frac 1 {\sqrt2} \log\left|\tan\left(\frac {3\pi} 8-\frac x 2\right)\right|+C}(**) $$

But I got a different result. Could you please explain the reason for this anomaly? Is it wrong to do a different replacement in step $(1)$? I think it shouldn't make any difference.

Further, could you please explain how to think we must be doing the replacement instead of using $1+\sin2x=(\sin x+\cos x)^2$ to solve this integral? I got this idea only after looking the solution.

*Using $\int \csc x dx=\log\left|\tan\left(x/2\right)\right|+C$

**Using $\int \sec x dx=\log\left|\tan\left(\frac \pi 4 +\frac x 2\right)\right|+C$

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    $\begingroup$ Are you sure that this is different ? $\endgroup$ – Yves Daoust Dec 18 '19 at 10:23
  • $\begingroup$ You might want to check signs, since $\frac 1 {\sqrt2} \log\left|\tan\left(\frac {3\pi} 8-\frac x 2\right)\right| = - \frac 1 {\sqrt2} \log\left|\tan\left(\frac \pi 8+\frac x 2\right)\right|$ $\endgroup$ – Henry Dec 18 '19 at 10:26
  • $\begingroup$ @YvesDaoust, I think yes. But it must be no. Is there anything wrong in my attempt? $\endgroup$ – Guru Vishnu Dec 18 '19 at 10:31
  • $\begingroup$ @Henry, Could you please explain how you transformed the final solution obtained by me to the negative of the solution given in the book? Even then why is there a difference in the signs of the final expression? $\endgroup$ – Guru Vishnu Dec 18 '19 at 10:33
  • $\begingroup$ My point is really that $\tan\left(\frac\pi 4+y\right)\, \tan\left(\frac\pi 4-y\right)=1$ so taking logarithms turns this into a sign change $\endgroup$ – Henry Dec 18 '19 at 10:38
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I think I prefer your first idea for its cuteness. You're only a step away, for you have that $$\sin x+\cos x=\sqrt 2\sin\left(x+\fracĪ€4\right).$$

Then you would only want to do something which is a constant multiple of the form $$\int\csc y\mathrm dy,$$ which is a standard integral.

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  • $\begingroup$ Thank you. You're right. I could have avoided this much trouble. If possible, could you please explain the reason for the inconsistency between the two methods used? $\endgroup$ – Guru Vishnu Dec 18 '19 at 10:40
  • $\begingroup$ @M.GuruVishnu The problem is with the second method. What you have is a constant multiple of $$\int\sec\left(x-\fracĪ€4\right)\mathrm d\left(x-\fracĪ€4\right)=\log\left|\tan\left(\frac x2+\fracĪ€8\right)\right|+C,$$ as before. If you wanted to do it your way then you should have had $$\color{red}{-}\int\sec\left(\fracĪ€4-x\right)\mathrm d\left(\fracĪ€4-x\right)$$ instead (note the sign change). $\endgroup$ – Allawonder Dec 18 '19 at 11:00

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