# How do I evaluate int(t +5) sin (t+5) dt?

Jan 31, 2015

The answer is: $- \left(t + 5\right) \cos \left(t + 5\right) + \sin \left(t + 5\right) + c$

This integral has to be done with the theorem of the integration by parts, that says:

$\int f \left(t\right) g ' \left(t\right) \mathrm{dt} = f \left(t\right) g \left(t\right) - \int g \left(t\right) f ' \left(t\right) \mathrm{dx}$

We can assume that $f \left(t\right) = t + 5$ and $g ' \left(t\right) \mathrm{dt} = \sin \left(t + 5\right) \mathrm{dt}$.

So:

$g \left(t\right) = \int \left(\sin \left(t + 5\right) \mathrm{dt}\right) = - \cos \left(t + 5\right)$,

$f ' \left(x\right) \mathrm{dx} = 1 \mathrm{dx}$.

Then:

$\int \left(t + 5\right) \sin \left(t + 5\right) \mathrm{dt} = \left(t + 5\right) \left(- \cos \left(t + 5\right)\right) - \int - \cos \left(t + 5\right) 1 \mathrm{dt} = - \left(t + 5\right) \cos \left(t + 5\right) + \sin \left(t + 5\right) + c$