Calculating $\int_0^{\pi/2} \frac{\ln(\sin x+\cos x)}{\sin x\cos x}\, dx$ How would you evaluate the following integral:
$$\int_0^{\pi/2} \frac{\ln(\sin x+\cos x)}{\sin x\cos x}\, dx$$ 
I can't see any helpful by parts or substitution ideas, e.g. $u=\sec x$ or $u=\csc x$ since the other integral that comes out of by parts don't evaluate nicely, so any help would be much appreciated. The integral should come out nicely as it’s part of a problem (at roughly first year undergraduate level) that says show a particular integral comes to something nice, which I’ve simplified to this.
 A: Rewrite the integral as: $$\int_0^\frac{\pi}{2}\frac{2\ln(\sin x+\cos x)}{2\sin x\cos x}dx=\int_0^\frac{\pi}{2}\frac{\ln(1+\sin (2x))}{\sin (2x)}dx=\int_0^\frac{\pi}{2}\frac{\ln (1+\sin x)}{\sin x}dx =\frac{\pi^2}{8}$$ For the last integral look here and take $a=1$.
A: I do not know if this hint is conclusive, but this is what I got:
\begin{align*}
\frac{\ln(\sin(x)+\cos(x))}{\sin(x)\cos(x)} & = \frac{\ln[\left(\tan(x) + 1\right)\cos(x)]}{\cos^{2}(x)\tan(x)}\\
& = \frac{\ln(\tan(x)+1)\times\sec^{2}(x)}{\tan(x)} + \frac{\ln(\cos(x))\times\sec^{2}(x)}{\tan(x)}
\end{align*}
On the one hand, the substitution $u = \tan(x) + 1$ results into
\begin{align*}
\int\frac{\ln(\tan(x)+1)\times\sec^{2}(x)}{\tan(x)}\mathrm{d}x = \int\frac{\ln(u)}{u-1}\mathrm{d}u
\end{align*}
On the other hand, since $\sec^{2}(x) = 1 + \tan^{2}(x)$, the same substitution results into
\begin{align*}
\int\frac{\ln(\cos(x))\times\sec^{2}(x)}{\tan(x)}\mathrm{d}x = \int\frac{\ln(2 - 2u + u^{2})^{-1/2}}{u - 1}\mathrm{d}u = \frac{1}{2}\int\frac{\ln(2-2u+u^{2})^{-1}}{u-1}\mathrm{d}u
\end{align*}
Thus the problem to be solved is equivalent to
\begin{align*}
\int_{0}^{\pi/2}\frac{\ln(\sin(x)+\cos(x))}{\sin(x)\cos(x)}\mathrm{d}x = 
\frac{1}{2}\int_{1}^{+\infty}\ln\left(\frac{u^{2}}{2-2u+u^{2}}\right)\frac{\mathrm{d}u}{1-u}
\end{align*}
A: Note that if you use Weierstrass substitution you get:
$$I=2\int_0^1\frac{(1+t^2)(\ln(1+2t-t^2)-\ln(1+t^2))}{2t(1-t^2)}dt$$
This may work
EDIT
note:
$$\sin(x)+\cos(x)=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$$
also, we know the rule:
$$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$
but when applied to the original integral it does not help us.
Writing our integral as:
$$I=\int_0^{\pi/2}\frac{\ln\left(\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\right)}{\sin(x)\cos(x)}dx$$
we can now apply the substitution:
$$u=x+\frac{\pi}{4}$$
to obtain:
$$I=2\int\limits_{\pi/4}^{3\pi/4}\frac{\ln\left(\sqrt{2}\sin(u)\right)}{(\sin u+\cos u)(\sin u-\cos u)}du=2\int\limits_{\pi/4}^{3\pi/4}\frac{\ln\left(\sqrt{2}\sin(u)\right)}{\sin^2(u)-\cos^2(u)}du$$
now we would like to substitute with the $\sqrt{2}\sin(u)$ but this doesn't work with the limits. So we are going to try using the substitution $v=\sqrt{2}\sin(u)$ and split the integral up into two parts.
$$I_1=2\int\limits_{\pi/4}^{\pi/2}\frac{\ln(\sqrt{2}\sin(u)}{\sin^2(u)-\cos^2(u)}du=\sqrt{2}\int_1^\sqrt{2}\frac{\ln(v)}{\frac{v^2}{2}-\frac{2-v^2}{2}}dv=\sqrt{2}\int_1^\sqrt{2}\frac{\ln(v)}{v^2-1}dv$$$$
=\sqrt{2}\int_1^\sqrt{2}\frac{\ln(v)}{2(v-1)}-\frac{\ln(v)}{2(v+1)}dv$$
and the same can be done for the second part.
