Checking the result of integration, $\int\sinh^3x\cosh xdx$ I integrated $\int\sinh^3x\cosh xdx$ in the following way:
\begin{align*}
\int\sinh^3x\cosh xdx ={}& \int\sinh^2x\sinh x\cosh xdx = \frac12\int\sinh^2x\sinh(2x)dx ={} \\
{}={}& \frac14\int(\cosh(2x)\sinh(2x)-\sinh(2x))dx ={} \\
{}={}& \frac18\int\sinh(4x)dx - \frac14\int\sinh(2x)dx ={} \\
{}={}& \frac{1}{32}\cosh(4x) - \frac18\cosh(2x) + c.
\end{align*}
The textbook answer is: $\frac14(\sinh x)^4 + c$
Are these two expressions equal or is there an error in my calculations? 
 A: Here is one way to show they are the same:
$$\frac14(\sinh^4x)=\frac14\left(\frac{(e^x-e^{-x})}{2}\right)^4\\=\frac{e^{4x}-4e^{2x}+6-4e^{-2x}+e^{-4x}}{64}\\=\frac{1}{32}\cosh4x-\frac18\cosh2x+\frac6{64}$$
A: Let us look at all equalities one by one:


*

*First equals: right.

*Second: right.

*Third equals: $\cosh(2x)=\cosh^2x+\sinh^2x=1+2\sinh^2x$, so $\sinh^2x=\frac{\cosh(2x)-1}{2}$, so OK.

*Fourth equals: split integral by linearity and $\sinh(2x)=2\sinh x\cosh x$, so OK.

*Fifth equals: um, the first integral is wrong! Differentiating it gives me $\frac{4}{32}\sinh(4x)$, so the $\cosh x$ in the integral was eaten up in integration! And the second term… that is right. Edit Seems that $\cosh x$ was out of place. So this passage becomes true.


As for how to get the correct result more easily, notice (cfr this comment) that the derivative of $\sinh$ is $\cosh$, so substituting $t=\sinh x$ will give you:
$$\int\sinh^3x\cosh xdx=\int t^3dt=\frac{t^4}{4}+c=\frac{\sinh^4x}{4}+c.$$
Which agrees with the textbook.
Let's see if they are the same:
\begin{align*}
\frac{\sinh^4x}{4}={}&\frac14(\sinh^2x)^2=\frac14\left(\frac{\cosh(2x)-1}{2}\right)^2=\frac{1}{16}(\cosh^2(2x)-2\cosh(2x)+1)={} \\
{}={}&\frac{1}{16}\left(\frac{\cosh(4x)+1}{2}-2\cosh(2x)+1\right)=\frac{1}{32}\cosh(4x)-\frac18\cosh(2x)+\frac{1}{16}+\frac{1}{32}.
\end{align*}
So the parts of those integrals without the $c$ are equal up to a constant. Hence, putting all constants together, the integrals are equal.
A: $$\int \sinh ^3\left(x\right)\cosh \left(x\right)dx$$
Apply Integral Substitution $\color{green}{u=\sinh \left(x\right)\quad \:du=\cosh \left(x\right)dx}$
Then
$\Rightarrow \:du=\cosh \left(x\right)dx$
$\Rightarrow \:dx= sech \left(x\right)du$
$$=\int \:u^3\cosh \left(x\right)sech \left(x\right)du=\int \:u^3du=\frac{u^{3+1}}{3+1}$$
So
$$\int \sinh ^3\left(x\right)\cosh \left(x\right)dx=\color{red}{\frac{\sinh ^4\left(x\right)}{4}+C}$$
