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Zarby
Integral using parts of t^(2) sin(t) I got it to -t^2cos(t) dt+2Integral tcos(t) dt How and why do I have to use parts again to solve this? Thanks
So you've gotten to the point \[\int t^2\sin(t) dt=-t^2\cos(t)+2\int t\cos(t) dt\]To finish solving this, you need to integrate \[\int t\cos(t) dt\] using parts in the same way that you did the first integral.
You have to use parts to solve it, because it's the best way of solving it after guessing.
Why only integrate that side?
What exactly do you mean?