?sin2xcos6xdx

Question

∫sin²xcos⁶xdx

Answer 1

Maybe these substitutions will work

 

sin²x= (1 - cos2x)/2

cos²x = (1 + cos2x)/2

 

(1/16) ∫ (1 - cos2x)(1 + cos2x)³ dx

 

multiply through and if you get a term such as cos²(2x), change it to (1 + cos(4x))/2

Answer 2

If the 2 and 6 are powers

∫sin²xcos⁶xdx

I would replace sin² with 1-cos² , then use U substitution

 

If the 2 and 6 are double angles

∫sin(2x)*cos(6x)dx

You'll have to use double angle formulas, and reduce the 6x also by using a triple and then a double angle formula. Use this site as a reference

http://www.mathreference.com/ca-int,daf.html

 

 

Edit: The comment below me is correct, I was thinking of differentiation as I was more concerned with properly parsing the question. I updated my responses with a workable process

 

 

I'd use this page as a good reference for possibilities.

http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithTrig.aspx