Evaluate ∫π/20xsinxcosxsin4x+cos4xdx Source : Putnam
By the property ∫a0f(x)dx=∫a0f(a−x)dx:
=∫π/20(π/2−x)sinxcosxsin4x+cos4xdx=π2∫π/20sinxcosxsin4x+cos4xdx−∫π/20xsinxcosxsin4x+cos4xdx
⟺∫π/20xsinxcosxsin4x+cos4xdx=π4∫π/20sinxcosxsin4x+cos4xdx
Now I'm stuck. WolframAlpha says the indefinite integral of sinxcosxsin4x+cos4x evaluates nicely to −12arctan(cos(2x)).
I already factored sin4x+cos4x into 1−(sin(2x)√2)2, but I don't know how to continue.. I suggest a substitution u=sin(2x)√2?
Could someone provide me a hint, or maybe an easier method I can refer to in the future?
Answer
1−sin2(2x)2=1+cos2(2x)2, and sinxcosx=sin(2x)2⇒∫sinxcosxcos4x+sin4xdx=∫−12d(cos(2x))1+cos2(2x)dx=−12arctan(cos(2x))+C
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