Question

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

Answer 1

The answer is: `-(t+5)cos(t+5)+sin(t+5)+c`

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

`intf(t)g'(t)dt=f(t)g(t)-intg(t)f'(t)dx`

We can assume that `f(t)=t+5` and `g'(t)dt=sin(t+5)dt`.

So:

`g(t)=int(sin(t+5)dt)=-cos(t+5)`,

`f'(x)dx=1dx`.

Then:

`int(t+5)sin(t+5)dt=(t+5)(-cos(t+5))-int-cos(t+5)1dt=-(t+5)cos(t+5)+sin(t+5)+c`