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Question

Obtain reduction formula for In=sinnx dx,n being a positive integer, n2 and hence deduce the value of sin4x dx

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Solution

In=sinnxdx=sinn1xsinxdx
now do integration by parts , we get In=sinn1x(cosx)+(n1)cos2xsinn2x=sinn1xcosx+(n1)sinn2x(1sin2x)
In=sinn1xcosx+(n1)In2(n1)In
nIn=(n1)In2sinn1xcosx
Now put n=4 , we get 4sin4xdx=3sin2xdxsin3xcosx=3(xsinxcosx2)sin3xcosx
Therefore sin4xdx=14(3(xsinxcosx2)sin3xcosx)

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