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
Mathematica does give the value of $I$. The question is how to get it by hand?
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$\begingroup$ What did Mathematica give? $\endgroup$– SzetoCommented Jul 2, 2019 at 6:50
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$\begingroup$ @Szeto Oh! It is a bit lengthy expression. $\endgroup$– Z AhmedCommented Jul 2, 2019 at 6:56
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$\begingroup$ @Szeto it gives $ (-1 - 3 e^4 + (1 + e^4) Log[1/2 (1 + e^4)])/(-1 + e^4)$ $\endgroup$– Z AhmedCommented Jul 2, 2019 at 7:02
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1 Answer
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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 $$I=\int_{0}^{1} \frac{\tanh^2 1-\tanh^2 t}{1-\tanh^2 1 \tanh^2 t} dt=(1+a^2)\int_{0}^{a} \frac{dz}{1-a^2z^2}-\int_{0}^{a} \frac{dz}{1-z^2}$$ $$=\frac{1+a^2}{a} \tanh^{-1} a^2-1.$$
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1$\begingroup$ Very snazzy (+1). much easier than I thought. $\endgroup$ Commented Jul 2, 2019 at 7:28
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$\begingroup$ @clathratus , Thanks. they say that beauty lies in the eyes of the beholder. $\endgroup$– Z AhmedCommented Jul 2, 2019 at 9:15