Show, by means of the substitution \(u=\cosh x\,\), that
\[
\int \frac{\sinh x}{\cosh 2x} \d x
= \frac 1{2\sqrt2}
\ln \left\vert \frac{\sqrt2 \cosh x - 1}{\sqrt2 \cosh x + 1 } \right\vert
+ C
\,.\]
Use a similar substitution to find an expression for
\[
\int \frac{\cosh x}{\cosh 2x} \d x
\,.\]
Using parts (i) and (ii) above, show that
\[
\int_0^1 \frac 1{1+u^4} \d u = \frac{\pi + 2\ln(\sqrt2 +1)}{4\sqrt2}\,.
\]
Solution:
\begin{align*}
&& \int \frac{\sinh x}{\cosh 2x} \d x &= \int \frac{\sinh x}{2\cosh^2 x -1} \d x \\
u = \cosh x, \d u = \sinh x \d x &&&= \int \frac{1}{2u^2 -1} \d u \\
&&&= \int\frac12 \left ( \frac{1}{\sqrt{2}u-1}-\frac{1}{\sqrt{2}u+1} \right) \d u \\
&&&= \frac1{2\sqrt{2}} \left (\ln (\sqrt{2}u-1) - \ln(\sqrt{2}u+1) \right) + C \\
&&&= \frac{1}{2\sqrt{2}} \ln \left ( \frac{\sqrt{2}\cosh x -1}{\sqrt{2}\cosh x +1} \right) + C
\end{align*}
\begin{align*}
&& \int \frac{\cosh x}{\cosh 2x} \d x &= \int \frac{\cosh x}{1+2\sinh^2 x} \d x \\
u = \sinh x && &= \int \frac{1}{1+2u^2} \d u \\
&&&=\frac{1}{\sqrt{2}} \tan^{-1} (\sqrt{2}u) + C \\
&&&= \frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2}\sinh x) + C
\end{align*}