Question

Please help. Integrate sinh^2(x) cosh^2(x) dx

Answer 1

Help pls.

Answer 2

For this one i would rewrite in exponential form
---> http://en.wikipedia.org/wiki/Hyperbolic_function#Standard_algebraic_expressions

changing the integral to
\[\int\limits \frac{ e^{-8x} -2 e^{-4x} +1}{16 e^{-4x}} dx\]
then use substitution
u = e^(-4x)
du = -4 e^(-4x) dx

\[-\frac{1}{64} \int\limits \frac{u^2 -2u +1}{u^2}\]
\[ = -\frac{1}{64}( u -2 \ln u - \frac{1}{u})\]
plugging back in and simplifying
\[-\frac{1}{64} (\frac{e^{-8x} -1}{e^{-4x}} + 8x)\]
distribute the neg and put back in hyperbolic form
\[= \frac{1}{32}( \sinh (4x) - 4x)\]

Answer 3

Thank you very much sir

Answer 4

Another suggestion: integrate by parts, setting
\[\begin{matrix}
u=\sinh x&&&dv=\sinh x\cosh^2x~dx\\
du=\cosh x~dx&&&v=\frac{1}{3}\cosh^3x
\end{matrix}\]
So
\[\begin{align*}\int\sinh^2x\cosh^2x~dx&=\frac{1}{3}\cosh^3x\sinh x-\frac{1}{3}\int\cosh^4x~dx\\\\
&=\frac{1}{3}\cosh^3x\sinh x-\frac{1}{3}\int\cosh^2x\cosh^2x~dx\\\\
&=\frac{1}{3}\cosh^3x\sinh x-\frac{1}{3}\int(1+\sinh^2x)\cosh^2x~dx\\\\
\frac{4}{3}\int\sinh^2x\cosh^2x~dx&=\frac{1}{3}\cosh^3x\sinh x-\frac{1}{3}\int \cosh^2x~dx\\\\
\int\sinh^2x\cosh^2x~dx&=\frac{1}{4}\cosh^3x\sinh x-\frac{1}{4}\int \frac{\cosh2x+1}{2}~dx\\\\
&=\cdots
\end{align*}\]