Question

integrate sinh^2(x) cosh^3 (x) dx

Answer 1

First thing's first: Hyperbolic Trig Identities (YAY!)
Recall:\[\cosh ^{2}(x)-\sinh ^{2}(x)=1\] Which form of this identity should we use to integrate this problem? Hint: Think of the u substitution you want to make later on.

Answer 2

so it will be like this
\[\int\limits \sinh^2 x \cosh x(\sinh^2 x+ 1)\] then let u=sinhx

Answer 3

Exactly. Continue, you're doing great!

Answer 4

\[du=\cosh x\]\[u^2 coshx(u^2 +1) \frac{ du }{ coshx }\] \[\int\limits u^2(u^2+1)\]

Answer 5

Don't forget your du at the end, and then you can just integrate and replace your u and you have your answer.

Answer 6

ok, thanks

Answer 7

I am glad to help, but did the entire problem. Nice work!

Answer 8

*but you did the entire problem.

Answer 9

i got the answer actually but since the answer is different from answer sheet, so i want confirmation.

Answer 10

You are my hyperbolic expert! :o

Answer 11

Well next time you can just post problem and your answer and someone will double check it if you want to skip me (or someone else) trying to explain all of the steps :P

Answer 12

noted, thanks again

Answer 13

\(\color{#000000 }{ \displaystyle \frac{u^5}{5} +\frac{u^3}{3}+C}\)

Recall that you did
\(\color{#000000 }{ \displaystyle u=\sinh(x)}\)

So, to sub back

Do you get a different answer?

Answer 14

What answer did you get?

Answer 15

yeah i got the same answer, but the sheet answer is \[-\frac{ sinhx }{ 8 } + \frac{ 1 }{ 4 }\sinh3x +\frac{ 1 }{ 80 }\sinh5x\]

Answer 16

It doesn't even have a C :( The horror!

Answer 17

lol, i know right

Answer 18

\(\color{#000000 }{ \displaystyle \frac{\sinh^5(x) }{5}+\frac{\sinh^3(x) }{3} +C }\)

that is what I seem to get - don't yet know if I can conviniently simplify this, if I could simplify this at all...

Answer 19

yeah that's the answer i got, dont know how on earth the sheet answer its so complex

Answer 20

\(\color{#000000 }{ \displaystyle 1-\sinh^2(x)\ne \cosh^2(x) ~~\forall x}\)

Answer 21

oh yeah i mislook that. thank you

Answer 22

\(\color{#000000 }{ \displaystyle \sinh^2(x)+\cosh^2(x)~~~~\square~~~~1}\)

\(\color{#000000 }{ \displaystyle \frac{1}{4}(e^x-e^{-x})^2+\frac{1}{4}(e^x+e^{-x})^2 ~~~~\square~~~~1}\)

\(\color{#000000 }{ \displaystyle (e^x-e^{-x})^2+(e^x+e^{-x})^2 ~~~~\square~~~~4}\)

\(\color{#000000 }{ \displaystyle (e^{2x}-2+e^{-2x})+(e^{2x}+2+e^{-2x}) ~~~~\square~~~~4}\)

\(\color{#000000 }{ \displaystyle 2e^{2x}+2e^{-2x} ~~~~\square~~~~4}\)

Answer 23

Again, using the definitions are very helpful, and the lesson is that hyperbolic functions don't (always) abide the same rules are regular trig functions do.

Answer 24

I don't know any hyperbolic properties; I haven't ever integrated them really. But, if I apply defs for sinh(x) and coh(x) then I can get some answer, although it might be painful and will definitely differ

Answer 25

\(\color{#000000 }{ \displaystyle \int \sinh^2(x) \cosh^3 (x) dx }\)

\(\color{#000000 }{ \displaystyle 2^5\int \left(e^{2x}+e^{-2x}-2\right)\left(e^{2x}+e^{-2x}+2\right)\left(e^{x}+e^{-x}\right) dx }\)

\(\color{#000000 }{ \displaystyle 32\int \left(\frac{e^{4x}+2e^{2x}-1}{e^{2x}}\right)\left(\frac{e^{4x}+2e^{2x}+1}{e^{2x}}\right)\left(\frac{e^{2x}+1}{e^x} \right) dx }\)

\(\color{#000000 }{ \displaystyle 32\int \frac{1}{e^{5x}}\left(e^{4x}+2e^{2x}-1\right)\left(e^{4x}+2e^{2x}+1\right)\left(e^{2x}+1 \right) dx }\)

I honestly would do this if I were on the test an didn't have the web to look for rules....

Answer 26

in our test the identities will be given, so one less burden to us, but thanks for the alternative way