Integrate by part please show your work thanks

please do step by step coz i'm new for this

∫ t²cos t dt

cos t = (sin t)' so to do integration by parts we can "move the derivative from the cos to the t^2" i.e. to end up with an integrand like 2tsin t and repeat the process to get -2cos t which we can easily evaluate.

In doing this we also get some extra terms which aren't integrals as follows (where we integrate between a and b):

∫t²cos t = b²sin b-a²sin a - ∫2tsin t repeating this step we get:
-∫2tsin t = 2b cos b- 2a cos a -2∫cos t = 2b cos b- 2a cos a -2sin b +2 sin a

Hence ∫t²cos t = b²sin b-a²sin a + 2b cos b- 2a cos a -2sin b +2 sin a

Altenatively if you are seeking the antiderivative i.e integrating between a and a variable s, we get:

∫t²cos t=s² sin s +2 s cos s -2 sin s +C

Where C is all the terms in a .

Remember that integration by parts is "derived" from the product rule for differentiation by integrating.

i.e. (uv)'=u'v+uv' and integrate to get : uv=∫u'v +∫uv' and rearrange to get ∫uv'=uv-∫u'v
In this case u(t)=t² and v(t)=sin t (so v'(t)=cos(t) )
Hope this helps.

{§ integral sign} , let u=t^2 then du=2tdt , dv=cost dt then v=sint , I=§t^2 cost dt = t^2 sint-2§tsint dt , in a similar we do the other integral {let u=t & dv=sint dt} , I=t^2 sint +2tcost -2sint

I don't think this has an integral.