# A integral with hyperbolic function

Evaluate : $$\int_{0}^{2{\rm arccosh\,} \pi }{\frac{\text{d}x}{1+\frac{{{\sinh }^{2}}x}{{{\pi }^{4}}}}}$$

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You could start getting rid of the double fraction by taking $\pi^4$ out to obtain $\pi^4 \cdot \int^{2 \arccos h \pi}_0 \frac{dx}{\pi^4+sinh^2x}$ – Bob Feb 1 '13 at 12:01

Expand $\sinh^2{x}$ into exponentials, and let $u=e^{-2 x}$. You end up with

$$2 \pi^4 \int_{e^{-4 \mathrm{arccosh}}(\pi)}^1 \frac{du}{1+2 (2 \pi^4-1) u + u^2}$$

This is a standard integral which results in

$$\frac{1}{2 \pi^2\sqrt{\pi^4-1}} \mathrm{arctanh}{\left ( \frac{2 (\pi^2-1)^{3/2}}{\pi (2\pi^2-1)} \right )}$$

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Set $u=\frac{\sinh(x)}{\pi^2}$. Then $$du=\frac{\cosh(x)}{\pi^2} dx$$ $$dx=\frac{\pi^2}{\cosh(x)} du=\frac{\pi^2}{\sqrt{1+u^2}} du$$

Your integral becomes $$\int_{0}^{2 \rm arccosh{\pi}} \frac{dx}{1+\frac{\sinh^2(x)}{\pi^4}}=\pi^2 \int_{0}^{a}\frac{du}{(1+u^2)^{3/2}}$$

where $a=\frac{\sinh(2 \rm arccosh{\pi})}{\pi^2}=\frac{2\sqrt{\pi^2-1}}{\pi}$, using classical hyperbolic relations. The last integral is well-known and shouldn't be too difficult.

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