Originally Posted by

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

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

This is what I get and how I got there.

$\displaystyle = x^2 Sinh(x) - \int {2xSinh(x)} \_dx \\ $

$\displaystyle = x^2 Sinh(x) - 2\int {xCosh(x)} \_dx \\ $

$\displaystyle = x^2 Sinh(x) - 2[xSinh(x) - \int {Sinh(x)\_dx} ] \\ $

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

$\displaystyle = (\ln 2)^2 Sinh(x) - 2[(\ln 2)Sinh(\ln 2) - Cosh(\ln 2)] \\$

$\displaystyle = \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?

$\displaystyle

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