What is #int sin^4xdx #?

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What is #int sin^4xdx #?

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Answer 1 · 2016-08-10T20:18:53

We know that

#sin^2x=1/2(1-cos2x)# and #cos^2 2x=1/2*(1+cos4x)#

Hence

#int sin^4xdx=int (1/2(1-cos2x))^2dx= 1/4*int(1-2cos2x+cos^2 2x)dx= 1/4*int1dx-2*1/4intcos2xdx+1/4intcos^2 2xdx= 1/4*x-1/4*sin2x+1/4*int((1+cos4x)/2)dx= 1/4*x-1/4*sin2x+1/8*x+1/32*sin(4x)+c#

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What is #int sin^4xdx #?

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