If $I_n$ is defined as $\int^1_0\sinh^nx$ show that $nI_n+(n-1)I_{n-2}=\cosh1\sinh^{n-1}1$

Question

I've tried evaluating the first three terms, so I have the results:

$I_1=\cosh1-1$

$I_2=\frac{1}{4}\sinh2-\frac{1}{2}$

$I_3=\frac{1}{12}\cosh3-\frac{3}{4}\cosh1+\frac{2}{3}$

These do satisfy the result.

I'm really not sure where to progress from here.

Answer 1

Integration by parts, I guess:

$$\begin{cases}u=\sinh^{n-1}x,\,&u=(n-1)\cosh x\sinh^{n-2}x\\{}\\v'=\sinh x,\,&v=\cosh x\end{cases}\;\;\;\implies\,(\text{Observe}\;\cosh^2x=1+\sinh^2x)$$

$$ I_n=\left.\int_0^1\sinh x\,dx=\cosh x\sin^{n-1}x\right|_0^1-(n-1)\int_0^1\overbrace{\left(\sinh^{n-2}x+\sinh^nx\right)}^{=\cosh^2x\sinh^{n-2}x}dx=$$

$$=\cosh1\sinh^{n-1}1-(n-1)\left(I_{n-2}+I_n\right)\implies$$$${}$$

$$nI_n=\cosh1\sinh^{n-1}1-(n-1)I_{n-2}$$