Simplification of integrals.

Question

When dealing with hyperbolic functions, simplifying: $$ 2\pi\int_{-a}^a a \cdot \cosh^2\left(\frac{x}{a}\right)dx $$ Yields: $$ 2\pi a\cdot \frac{a}{2}(2+\sinh 2) $$

How is this possible?

Answer 1

$$ \cosh(2u) = 2\cosh^2 u -1 $$ Thus we can let your integral be $$ 2\pi\int_{-a}^{a} a\cosh^2\left(x/a\right) dx= 2\pi a^2\int_{-1}^{1} \cosh^2(u)du $$ Then using the identity I provide above you can proceed.

Answer 2

Rewrite this integral using $x=at$: $$I = 2\pi a\cdot a \int_{-1}^1 \cosh^2 t dt.$$ Now we use definition of $\cosh$: $$\cosh^2 t = \left(\frac{e^t + e^{-t}}{2}\right)^2 = \frac14(e^{2t} + 2 + e^{-2t}),$$ and then \begin{align} I &= 2\pi a^2 \cdot\frac14 \int_{-1}^1 (e^{2t} + 2 + e^{-2t})\, dt = 2\pi a^2 \cdot\frac14 \left.\left(\frac12e^{2t} + 2t - \frac12 e^{-2t}\right)\right|_{t=-1}^{t=1} =\\&= 2\pi a^2 \cdot\frac14 \left(\frac12 e^2 + 2 - \frac12e^{-2} - \frac12e^{-2} - (-2) + \frac12 e^2\right)=\\ &=\frac12\pi a^2 (e^2 - e^{-2} + 4) = \pi a^2 \left(\frac{e^2 - e^{-2}}{2} + 2\right)=\pi a^2 (2 + \sinh 2) \end{align}