1. integral x^2 Sinh(x)dx

Hi again. Trying to solve this by parts. Question is evaluate

$\int\limits_0^{\ln 2} {x^2 } Sinh(x)\_dx$

This is what I get and how I got there.

$= x^2 Cosh(x) - \int {2xSinh(x)} \_dx \\$
$= x^2 Cosh(x) - 2\int {xCosh(x)} \_dx \\$
$= x^2 Cosh(x) - 2[xSinh(x) - \int {Sinh(x)\_dx} ] \\$
$= x^2 Cosh(x) - 2[xSinh(x) - Cosh(x)] \\$
$= (\ln 2)^2 Cosh(x) - 2[(\ln 2)Sinh(\ln 2) - Cosh(\ln 2)] \\$
$= \frac{5}{4}(\ln 2)^2 - \frac{3}{2}(\ln 2) + \frac{5}{2} \\$

However the answer on the sheet is as below. What am I doing wrong?
$
= \frac{5}{4}(\ln 2)^2 - \frac{3}{2}(\ln 2) + \frac{1}{2}$

2. Originally Posted by Craka
Hi again. Trying to solve this by parts. Question is evaluate

$\int\limits_0^{\ln 2} {x^2 } Sinh(x)\_dx$

This is what I get and how I got there.

$= x^2 Sinh(x) - \int {2xSinh(x)} \_dx \\$
$= x^2 Sinh(x) - 2\int {xCosh(x)} \_dx \\$
$= x^2 Sinh(x) - 2[xSinh(x) - \int {Sinh(x)\_dx} ] \\$
$= x^2 Sinh(x) - 2[xSinh(x) - Cosh(x)] \\$
$= (\ln 2)^2 Sinh(x) - 2[(\ln 2)Sinh(\ln 2) - Cosh(\ln 2)] \\$
$= \frac{5}{4}(\ln 2)^2 - \frac{3}{2}(\ln 2) + \frac{5}{2} \\$

However the answer on the sheet is as below. What am I doing wrong?
$
= \frac{5}{4}(\ln 2)^2 - \frac{3}{2}(\ln 2) + \frac{1}{2}$
I think your first mistake would be

I assumed you let $u=x^2$ and $dv=\sinh(x)dx$

then $du=2xdx$ and $v=\color{red}{\cosh(x)}$

unless I am mistaken

3. Thanks for you reply, but sorry I don't follow. I did let u=x^2 dv= Sinh(x)
du= 2x dx and v= Cosh(x)
Are you saying should be v= -Cosh(x) ?

4. Originally Posted by Craka
Thanks for you reply, but sorry I don't follow. I did let u=x^2 dv= Sinh(x)
du= 2x dx and v= Cosh(x)
Are you saying should be v= -Cosh(x) ?
Oh, I am sorry that was a typo, yes you are correct in that v=cosh(x)

But isnt the formula for parts $\int{udv}=uv-\int{vdu}$

So dont you have

$x^2\sinh(x)$ which is equal to $udv$?

5. Sorry I made a type in posting my working I have now corrected it, but question still remains on what I have done wrong.

6. Originally Posted by Craka
Sorry I made a type in posting my working I have now corrected it, but question still remains on what I have done wrong.
Originally Posted by Craka
Hi again. Trying to solve this by parts. Question is evaluate

$\int\limits_0^{\ln 2} {x^2 } Sinh(x)\_dx$

This is what I get and how I got there.

$= x^2 Cosh(x) - \int {2xSinh(x)} \_dx \\$
$= x^2 Cosh(x) - 2\int {xCosh(x)} \_dx \\$
$= x^2 Cosh(x) - 2[xSinh(x) - \int {Sinh(x)\_dx} ] \\$
$= x^2 Cosh(x) - 2[xSinh(x) - Cosh(x)] \\$
$= (\ln 2)^2 Cosh(x) - 2[(\ln 2)Sinh(\ln 2) - Cosh(\ln 2)] \\$
$= \frac{5}{4}(\ln 2)^2 - \frac{3}{2}(\ln 2) + \frac{5}{2} \\$

However the answer on the sheet is as below. What am I doing wrong?
$
= \frac{5}{4}(\ln 2)^2 - \frac{3}{2}(\ln 2) + \frac{1}{2}$
I'm gonna take a shot at this..I haven't dealt with hyperbolic trig in a while..

I believe you have it right, up to here:

$= x^2 Cosh(x) - 2[xSinh(x) - Cosh(x)] \\$

Apply the FTC, we get:

$=\left[ (\ln(2))^2 \cosh(\ln(2)) - 2[\ln(2)\sinh(\ln(2)) - \cosh(\ln(2))\right] {\color{red}-\left[-2(-\cosh(0))\right]}
$

$=\frac{5}{4}(\ln(2))^2-\frac{3}{2}\ln(2)+\frac{5}{2}-2$
$=\color{red}\boxed{\frac{5}{4}(\ln(2))^2-\frac{3}{2}\ln(2)+\frac{1}{2}}$

Hope this makes sense!

7. Where/why did the -Cosh(0) come from?
Are you applying the FTC thereom of f(b)-f(a) and if so would you not substitute 0 through all of the expression to evaluate?

8. Originally Posted by Craka
Where did $=\left[ (\ln(2))^2 \cosh(\ln(2)) - 2[\ln(2)\sinh(\ln(2)) - \cosh(\ln(2))\right] {\color{red}-\left[-2(-\cosh(0))\right]}$

come from. If doing that wouldn't you sub zero in for the whol expression? I take it as you are doing f(b)-f(a) ??
Yes, you would substitute into the whole thing, but note that $\sinh(0)=0$. Thus, all those terms drop out except for the last one.

Yes, I was doing f(b)-f(a).

9. Sorry that is exactly what you are doing. Just realised of my stupidity.