# Thread: Substition and Integration By Parts

1. ## Substition and Integration By Parts

Evaluate $\displaystyle \int2cos(ln t)dt$ by using a substitution prior to integration by parts.

Thanks for any help!

2. Try u = ln(t). Then t = exp(u).
du = dt/t so dt = exp(u)du.

Now you've got an integral with
cos(u)exp(u)du.

Remember cos(u) = {exp(iu)+exp(-iu)}/2 ?

Now it's all exponentials.

3. $\displaystyle 2* \int{cos(ln(t))dt}$

$\displaystyle u=ln(t)$ and $\displaystyle du=\frac{1}{t}$

$\displaystyle 2* \int{cos(u)(\frac{1}{t}du)}$

Take your u equation and solve for t now.

$\displaystyle t={e}^{-u}$

$\displaystyle 2* \int{e^{-u}*cos(u)du}$

$\displaystyle u=cos(u)$
$\displaystyle du=-sin(u)$
$\displaystyle dv={e}^{-u}$
$\displaystyle v=-e^{-u}$

Solve and then back substitute

$\displaystyle uv- \int (vdu)$

4. Use the substitution u=ln(t) and then use the formula

$\displaystyle \int e^{\alpha u} \cos(\beta u) du = \frac{e^{\alpha u} (\alpha \cos(\beta u)+\beta \sin(\beta u))}{\alpha^2+\beta^2}+C$

which you can calculate yourself, instead of memorizing. (I personally prefer the former because you don't need it that much.)