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sin3xcos3xdx is equal to 

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a
132−32cos⁡2x+16cos⁡6x+C
b
116−32cos⁡2x+16cos⁡6x+C
c
164−32cos⁡2x+16cos⁡6x+C
d
None of the above

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detailed solution

Correct option is A

∫sin3⁡xcos3⁡xdx=18∫(2sin⁡xcos⁡x)3dx                                 =18∫sin3⁡2xdx=18∫3sin⁡2x−sin⁡6x4dx                                 =132∫(3sin⁡2x−sin⁡6x)dx                                 =132−32cos⁡2x+16cos⁡6x+C


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