Friday, 15 April 2016

calculus - The Intergral of sin4x without using reduction formula




So I've been trying to compute sin4(x)dx and everywhere they use the reduction formula which we haven't learned yet so I've been wondering if theres another way to do it? Thanks in advance.


Answer



Performing integration by parts,




x0sin2tdt=[costsint]x0+x0cos2tdt=cosxsinx+x0(1sin2t)dt=cosxsinx+x01dtx0sin2tdt=cosxsinx+xx0sin2tdt



Therefore,



x0sin2tdt=12cosxsinx+12x




x0sin4tdt=x0(1cos2)sin2tdt=x0sin2tdtx0cos2tsin2tdt=12cosxsinx+12xx0cos2tsin2tdt



Since, for t real,



sin(2t)=2sintcost




then,



x0sin4tdt=12cosxsinx+12x14x0sin2(2t)dt



In the latter integral perform the change of variable y=2t,



x0sin4tdt=12cosxsinx+12x182x0sin2(y)dy=12cosxsinx+12x18(12cos(2x)sin(2x)+12×2x)=14sin(2x)+12x+132sin(4x)18x=14sin(2x)+38x+132sin(4x)



Therefore,



sin4xdx=38x+132sin(4x)14sin(2x)+C



(C a real constant)



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