# The integral $\int_{0}^{\pi/2} \sin (2x) ~ \tanh\left(2 \sin^2 x\right)~ \tanh\left(2\cos^2 x\right) ~dx$

Two integrals akin to $$I=\int_{0}^{\pi/2} \sin 2x~ \tanh(2 \sin^2 x) ~ \tanh(2\cos^2 x) ~dx$$ have been found interesting earlier

The integral $\int_{0}^{\pi/2} \sin 2\theta ~ \mbox{erf}(\sin \theta)~ \mbox{erf}(\cos \theta)~d\theta=e^{-1}$

Mathematica does give the value of $$I$$. The question is how to get it by hand?

• What did Mathematica give? Commented Jul 2, 2019 at 6:50
• @Szeto Oh! It is a bit lengthy expression. Commented Jul 2, 2019 at 6:56
• @Szeto it gives $(-1 - 3 e^4 + (1 + e^4) Log[1/2 (1 + e^4)])/(-1 + e^4)$ Commented Jul 2, 2019 at 7:02

Use $$\cos 2x=t$$, then $$I=\int_{0}^{\pi/2} \sin 2x \tanh(2 \sin^2 x) \tanh(2\cos^2 x) ~dx=\frac{1}{2}\int_{-1}^{1} \tanh(1-t) \tanh(1+t) dt.$$ Next, we use $$\tanh 1=a$$ and $$\tanh t =z$$, we get $$I=\int_{0}^{1} \frac{\tanh^2 1-\tanh^2 t}{1-\tanh^2 1 \tanh^2 t} dt=(1+a^2)\int_{0}^{a} \frac{dz}{1-a^2z^2}-\int_{0}^{a} \frac{dz}{1-z^2}$$ $$=\frac{1+a^2}{a} \tanh^{-1} a^2-1.$$