Hyperbolic trig integral (i know i could expand it but there must be another way)
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Hyperbolic trig integral (i know i could expand it but there must be another way)

[From: ] [author: ] [Date: 11-04-25] [Hit: ]
int(cosh(x)u)du however, i tried to do this by parts, but you cant.just FYI the book doesnt tend to tidy up all that much.........
okay, integral( (cosh(x)^2)(sinh(x)^3 ) dx
that's
the integral of cosh(x) squared times sinh(x) cubed, sorry about the lack of "math font"
if posible i'd like to be pointed in the right direction, or workings, i've played about for about 45 minutes using varous methods,

the closest i got with it as is is using the subsitution of u = sinh(x)^2 this got rid of a cosh and all the sinh's by giving
int(cosh(x)u)du however, i tried to do this by parts, but you can't.

the answer i'm heading for (back of book):

(1/5) (cosh(x)^5) - (1/3) (cosh(x)^3) + c

just FYI the book doesn't tend to tidy up all that much....so that's probably a result of their method
I've tried playing around with what i am integrating, and i can get no form that doesn't have the same problems, a recursive parts method sollution of some sort.

my fallback method is some expanding of the integral to get cosh and sinh into there e^x forms....but this is...probably quite silly.

Alec

Lastly and again:
this isn't my homework, i'm self teaching myself the MEI FP2 module, I don't want method, reading maths here is hard (method is allowed) I want a nudge in the right direction, somewhere that works. because my playing has got me no-where, I really want to know how, and i cannot get it right!

-
∫cosh^2(x) sinh^3(x) dx

= ∫cosh^2(x) sinh^2(x) sin h(x) dx

use the fact that cosh^2(x) - sinh^2(x) = 1 ==> sinh^2(x) = cosh^2(x) - 1

= ∫cosh^2(x) (cosh^2(x) - 1 ) sinh(x) dx

let cosh(x) = u
sinh(x) dx = du

∫u^2 ( u^2 - 1 ) du

= ∫[u^4 - u^2 ] du

= (1/5)u^5 - (1/3)u^3 + C

back substitute u = cosh(x)

= (1/5)cosh^5(x) - (1/3)cosh^3(x) + C
1
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