$$\int_0^{\ln2} \sinh^4x\,dx$$

So I've rewritten it like:

$$\int_0^{\ln2} (\sinh^2x\cosh^2x-\sinh^2x)\,dx$$

So it seems I have to reduce the power here, do formulas like

$$\mathrm{sin}^2x = \frac{\mathrm{cos}(2x)-1}{2}$$

work for hyperbolic functions normally, or is there another way?

  • 1
    $\begingroup$ Yes, there are analogous formulas for the hyperbolic trig functions but you need to be careful about the signs since they are not always the same as with the regular trig functions. I find it easiest to just re-derive these when I need them. For example, we see $$\sinh^2(x) = \frac{e^{2x} - 2 + e^{-2x}}{4} = \frac{2\cosh(2x) - 2}{4}= \frac{\cosh(2x) - 1}{2}.$$ Likewise you will find that $$\cosh^2(x) = \frac{\cosh(2x) + 1}{2}$$ $\endgroup$ – User8128 Apr 19 '17 at 15:25

HINT: use that $$\sinh^4(x)=\left(\frac{1}{2}\left(e^x-e^{-x}\right)\right)^4$$


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