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

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Answer 1 · 2015-01-31T14:53:17

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:

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

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