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Question

Integrate the function 1sin3xsin(x+α) w.r.t x

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Solution

Consider sin3xsin(x+α)=sin3x[sinxcosα+cosxsinα] using sin(A+B)=sinAcosB+cosAsinB
=sin4xcosα+cosxsin3xsinα
=sin4xcosα+cosxsin3xsinα×sinxsinx
=sin4x[cosα+cosxsinαsinx]
=sin4x[cosα+cotxsinα]
Hence sin3xsin(x+α)=sin4x[cosα+cotxsinα] .......(1)
Now dxsin3xsin(x+α)=dxsin4x[cosα+cotxsinα]
=dxsin4x[cosα+cotxsinα]
=1sin2x×1cosα+cotx.sinαdx
Let cosα+cotxsinα=t
dt=csc2xsinαdx
dx=sin2xdtsinαdt
1sin2x×1cosα+cotx.sinαdx=1sin2xt×1sinα×sin2xdt
=1sinαdtt
=1sinα(2t+c)
Put t=cosα+cotxsinα
=2sinα(2cosα+cotxsinα+c)

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