Evaluate the integral. Use Integration by Parts.?

Question

The integral of t sin2t dt.

Please show your work...Thanks. :)

Answer 1

The integration by parts formula is:
integral(f'g) = fg - integral(fg').
If we make f' = sin2t and g = t, then that second integral will only involve sin, and it should make everything easier.
So, that means f = -0.5cos2t, and g = t. Then g' = 1, so we get:
integral(t sin 2t) = -0.5tcos2t - integral(-0.5cos2t * 1)
That last integral now just becomes -0.25sin2t.
So the answer is -0.5tcos 2t + 0.25sin2t (+c).

Answer 2

To integrate this by parts, let
u = t dv = sin 2t dt
du = dt v = -1/2 cos 2t
Then the original integral is uv - int(vdu)
= -1/2 * t cos 2t + 1/2 int cos 2t dt
= -1/2 * t cos 2t + 1/4* sin 2t + C
It seems the previous post has a sign error.