Question

how do I find:
The integral of sinh^2(x)cosh^2(x)

Answer 1

wish i knew hypers better ....

Answer 2

Gimme a sec, this is going to take a minute to get all the steps down...

Answer 3

Try changing to one or the other
\[\cosh ^{2}x=1-\sinh ^{2}x\]

Answer 4

yes but we're still stuck with powers, and you can't substitute u = ...

Answer 5

you're probably right, but its 3am here, and i can't think straight :P

Answer 6

Yeah I think that is it because now your integral is elementary\[\int\limits_{}^{} \sinh ^{2}x-\sinh ^{4}x dx\]

Answer 7

It is no longer multiplication it is subtraction.

Answer 8

\[\int\limits (\sin^2h)(\cos^2h) \rightarrow \int\limits \cos^2h(x) \sin^2h(x) dx \rightarrow \frac{1}{4} \sin^3h(x) \cosh(x)+1/4 \int\limits \sin^2h(x) dx\]\[\rightarrow \frac{1}{4} \sin^3h(x) \cosh(x)+\frac{1}{4} \int\limits (\frac{1}{2} \cosh(2 x)-\frac{1}{2}) dx \]\[\rightarrow\frac{1}{4} \sin^3h(x) \cosh(x)+\frac{1}{4} \int\limits -\frac{1}{2} dx+\frac{1}{8} \int\limits \cosh(2 x) dx\]\[\rightarrow\frac{1}{16} \int\limits \cosh(u) du+\frac{1}{4} \sin^3h(x) \cosh(x)+\frac{1}{4} \int\limits -\frac{1}{2}dx\]\[\rightarrow\frac{1}{16} \int\limits \cosh(u) du-\frac{x}{8}+\frac{1}{4} \sin^3h(x) \cosh(x)\]\[= \frac{\sinh(u)}{16}-\frac{x}{8}+\frac{1}{4} \sin^3(x) \cosh(x)+c\rightarrow \frac{1}{32} (\sinh(4 x)-4 x)+c\]

Answer 9

that makes sense :) thank you for your help!

Answer 10

Phew, CraigWilliam, admire your worth ethic. Preed23 after some sleep, compare the two to see if it is the same.
\[(1/3)\sinh ^{3}x \cosh x-(1/5)\sinh ^{5}x \cosh x\]

Answer 11

:D

Answer 12

haha yess :P thanks aswell

Answer 13

I think that's right, I just went over it again on my white board...

Answer 14

Now your next assignment, break down your sinh of 4 angles and see if it is equal to mind. :)