How do you integrate: ∫π20sin(3x)cos(2x)dx ?

Answer 1

This is a hard problem to answer in this format. It took me a page on paper. Here are the key points:

1) Pull out the 20π. That is just a constant.

2) Use a trig substitution: sin(3x) = 3sin(x) - 4 sin³(x) and cos(2x) = 1 - 2sin²(x)

3) Multiply out the resulting two binomials to get 20π∫8sin⁵(x) -10sin³(x) + 3sin(x) dx

4) Break this into 3 separate integrals. For the first 160π∫sin⁵(x), break it into 160π∫sin⁴(x)sin(x) and then substitute in the trig identity [1-cos²(x)]² for sin⁴(x) then use a u substitution u = cos(x). The result after you are all completed is 160π[-cos(x) +2/3cos³(x) - 1/5cos⁵(x)]

5) For the second integral -200π∫sin³(x)dx, repeat the process above and you ultimately get: -200π[-cos(x) +1/3cos³(x)

6) The third integral easily solves to -60πcos(x)

7) Put them all together and combine like terms and you get: -20πcos(x) +40πcos³(x) - 160/5πcos⁵(x) + C

Answer 2

Note sin x cos y = 1/2 [sin (x + y) + sin (x − y)]

sin(3x) cos(2x) = 1/2 [sin(5x)+ sinx]

∫π20sin(3x)cos(2x)dx

= 20 π ∫(1/2)[sin(5x)+sinx]dx

= 10π ∫[sin(5x)+sinx]dx

= 10 π [(-1/5) cos(5x) + (-1)cos(x)] + C

=-2 π [cos(5x) + 5cos(x)] + C

where C is any constant