Question

integration of sin^3 2x dx


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Answer 1

Step-by-step explanation:

∫sin32xdx=∫(2sinxcosx)3dx=8∫sin3xcos3xdx∫sin3⁡2xdx=∫(2sin⁡xcos⁡x)3dx=8∫sin3⁡xcos3⁡xdx

Now, we can proceed in two ways.

8∫sin3xcos3xdx=8∫sin3x(1−sin2x)cosxdx=8∫sin3x−sin5xd(sinx)=2sin4x−43sin6x+C8∫sin3⁡xcos3⁡xdx=8∫sin3⁡x(1−sin2⁡x)cos⁡xdx=8∫sin3⁡x−sin5⁡xd(sin⁡x)=2sin4⁡x−43sin6⁡x+C

or

8∫sin3xcos3xdx=8∫(1−cos2x)cos3xsinxdx=8∫cos5x−cos3xd(cosx)