Question

Integrate: 2x cosh(x) dx, upper limit=ln(2), lower limit=0

Answer 1

Is this it?
\[\int\limits_{0}^{\ln 2}2x.\cosh x.dx\]

Answer 2

You have in the integrand 2x*cosh(x)
But you know that cosh(x)=(e^x + e^-x)/2=(e^x)/2 + (e^-x)/2
So you have in the integrand (what's inside the integral):
2x[(e^x)/2 + (e^-x)/2]=xe^x + xe^-x
First term Second Term
To integrate the first term xe^x you use integration by parts
Let u=x ===> du=dx
Let dv=(e^x) dx ====> v=integral of e^x so v=e^x

Plug this into your formula integral of vdu= uv- integral vdu
so integral of xe^x= xe^x - integral e^x dx
so integral of xe^x=xe^x -e^x PART A SOLUTION

Now to integrate the second term you use integration by parts again,
it will be very similar. Second term is xe^-x
Integral of xe^-x dx= uv - integral vdu
Let u=x ====> du=dx
Let dv=e^-x dx =====> v=integral of e^-x dx, so v=-e^-x

Thus integral of xe^-x dx=x (-e^-x) - integral -e^-x dx
=-xe^-x-e^-x PART B SOLUTION

So your answer is PART A SOLUTION + PART B SOLUTION evaluated from 0 to ln2
ANSWER:
= [ln2e^(ln2) -e^(ln2) -ln2e^(-ln2)-e^(-ln2)] - [0e^0 -e^0 -0e^-0 -e^-0]
Now simplify.

Answer 3

Thanks mashe, that helped a lot.
kropot: yes, that's the question.

Answer 4

I haven't integrated hyperbolic functions in a long time (but I simply used what it is in exponential form) there might be an easier way (IDK maybe there isn't)...please check if I made any mistakes...but this is pretty much the strategy. I suggest as I suggest to anyone i provide help to try it yourself after understanding the steps you have to take and see if you arrive at the correct answer.

Answer 5

Oh and Of course you are welcome! More than happy to help :-)

Answer 6

Hey mashe, I think there may be a slightly shorter way, cosh(x) simply integrates to sinh(x)... is this correct?

Answer 7

HAHHAHA YES!!! AWESOME :-)

Answer 8

I always take the longer route I like the extra work, seriously all my teachers tell me this...

Answer 9

good :) but in this question they give you some extra info (hints): cosh(ln(2))=5/4, sinh(ln(2))=3/4

Answer 10

Okay awesome here it goes:
So then integral of 2x coshx dx= 2* Integral x coshx
Let u=x =====> du=dx
Let dv=coshx dx======> v=sinhx
So we have
2* Integral x coshx=2[ uv - integral vdu]
=2 [xsinhx- integral sinhx dx] the terms in the parenthesis are
evaluated from x=0 to x=ln2
=2[xsinhx-coshx]
=2 {[ ln2 sinh(ln2) - cosh(ln2)]- [0sinh(0) -cosh(0)] }
= 2{[ln2 (3/4) -(5/4)]-[0 sinh(0)-cosh(0)]}

Answer 11

sinh(0)=0 and cosh(0)=1 simplify :-)

Answer 12

Which then equals \[(3(\ln (2))/2 - 1/2\]
Is this correct?

Answer 13

It would have to be correct haha I just looked up the answer and that's the right one! Thankyou mashe!

Answer 14

You are welcome Saltin8R happy to help and better when it is the easier way or the most appropriate way hahaha