Evaluate $\displaystyle \int2cos(ln t)dt$ by using a substitution prior to integration by parts.
Thanks for any help!
$\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
$\displaystyle uv- \int (vdu)$
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.)