Originally Posted by

**MechEng** Ah, I did consider the substitution you recommended. However, the section I am working on is trigonometric substitution. As such, I feel it would be best to stick with the trig substitution here.

So, if I am understanding this correctly... if I get my anti-derivative in terms of x, then there is no need to calculate new terminals? In this case my answer would simply be $\displaystyle -\frac{5}{18}-\frac{\sqrt{13}}{3}$ ? Mr F says: The second term is not correct. Take more care with the arithmetic.

What is the purpose of finding new terminals in this case? Mr F says: None.

I got my new terminals as shown below:

$\displaystyle x=1$

$\displaystyle \theta=x$

$\displaystyle 1=\frac{3}{2}tan\theta$

$\displaystyle tan\theta=\frac{2}{3}$

$\displaystyle \theta=tan^{-1}(\frac{2}{3})$

AND

$\displaystyle x=2$

$\displaystyle \theta=x$

$\displaystyle 2=\frac{3}{2}tan\theta$

$\displaystyle tan\theta=\frac{4}{3}$

$\displaystyle \theta=tan^{-1}(\frac{4}{3})$

Ah... does this look a little more correct? Mr F says: Yes.