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