# Thread: Definite integral of hyperbolic function

1. ## Definite integral of hyperbolic function

If $\displaystyle \displaystyle I_n=\int^1_0\frac{\sinh^{2n}x}{\cosh x} dx$, prove that

$\displaystyle \displaystyle I_{n+1}+I_n=\frac{\sinh^{2n+1} (1)}{2n+1}$

Show that $\displaystyle I_0=-\frac{\pi}{2}+2\arctan e$ and evaluate $\displaystyle I_3$.

My problem is showing that $\displaystyle I_0=-\frac{\pi}{2}+2\arctan e$.
$\displaystyle \displaystyle I_0=\int^1_0\frac{\sinh^0 x}{\cosh x}dx$
$\displaystyle \displaystyle =\int^1_0\frac{1}{\cosh x} dx$
$\displaystyle \displaystyle =\left[\frac{1}{\sinh x} \ln|\cosh x|\right]^1_0$
Problem is after substituting 1 and 0, i still can't get $\displaystyle -\frac{\pi}{2}+2\arctan e$
Thanks!

2. $\displaystyle cosh(x) = \frac{e^x + e^{-x}}{2}$

$\displaystyle I_o = \int^1_0\frac{2}{e^x+e^{-}x}dx$

$\displaystyle I_o = \int^1_0\frac{2e^x}{e^{2x} + 1}dx$

Substitute e^x = t and solve the integration.

3. $\displaystyle cosh(x) = \frac{e^x + e^{-x}}{2}$

$\displaystyle I_o = \int^1_0\frac{2}{e^x+e^{-x}}dx$

$\displaystyle I_o = \int^1_0\frac{2e^x}{e^{2x} + 1}dx$

Substitute e^x = t and solve the integration.