Part of an integration Question?

Question

Integrate the the following

3sin2t - 2cos3t

Answer 1

This is just an application of reversing the chain rule.

You can use substitution to make the process easier to see and do.

∫ [ 3·sin(2·t) - 2·cos(3·t) ] dt

= 3·∫ sin(2·t) dt - 2·∫cos(3·t) dt

Let u = 2·t

Then du = 2 dt

And dt = du / 2

Let v = 3·t

Then dv = 3 dt

And dt = dv / 3

Substitute in u, du, v, and dv:

→ 3·∫ sin(u)du/2 - 2·∫cos(v)dv/3

= -3·cos(u)/2 - 2·sin(v)/3 + C

Reverse substitute:

= -3·cos(2·t)/2 - 2·sin(3·t)/3 + C

= -(3/2)·cos(2·t) - (2/3)·sin(3·t) + C

Answer 2

Integral[(3sin2t - 2cos3t) dt] = Integral[3sin2t dt] - Integral[2cos3t dt] = (-3/2)cos2t - (2/3)sin3t + C.  

Answer 3

-3/2  *   cos2t  -  2/3 . sin3t  + C