Solution assignment 05 Integration by parts

Calculate:

$\displaystyle\int{\sin(\ln(t))}dt$

We apply the substitution method:

$u=\ln(t)$

and calculate:

$du=\displaystyle\frac{1}{t}dt$

or (because $t=e^u$ ):

$dt=e^{-u}du$

Then we get:

$\displaystyle\int{\sin(u)e^{-u}}du$

We have calculated this integral in assignment 2 of this series. After some more calculations we get the right answer.