Hyperbolic question tough

Let In = integral of cosh^n (x) dx. Show nIn = sinhxcosh^(n-1) (x) + (n-1)I n-2
Hence show the integral from 0 to ln 2 of cosh^4 (x) dx = 3/8 (245/128 +ln 2)

Really not sure how to start or approach this problem. Any help appreciated

Since you're expressing I(n) in terms of I(n-2) you're looking to set up a reduction formula, which usually implies integration by parts.

The fact that there's a difference of 2 between n and n-2 suggests you either need to do the IBP twice, or you need to split out a square factor. In this case it's the latter.

Use the obvious identity on cosh^2(x), split it up, then look at how you could do IBP on part of it.

PS: Your posts would be considerably more readable if you start using LaTeX.

Okay, after doing a bit of research on what integration by reduction actually is (I haven't learnt it) I think I get what you mean.

however in this example, where it's simply cos instead of cosh. I understand why the (n-1) bit is factored out, but why does it goes from - the integral
to
+(n-1) x the integral.

why isn't it - (n-1) x the integral?

Okay, after doing a bit of research on what integration by reduction actually is (I haven't learnt it) I think I get what you mean.

however in this example, where it's simply cos instead of cosh. I understand why the (n-1) bit is factored out, but why does it goes from - the integral
to
+(n-1) x the integral.

why isn't it - (n-1) x the integral?