Answer :


let

Put t=cosx


dt=-sinxdx





Putting


A(1+t)(3+2t)+B(1-t)(3+2t)+C(1+t)(1-t)=1


Now Putting 1+t=0


t=-1


A(0)+B(2)(3-2)+C(0)=1



Now Putting 1-t=0


t=1


A(2)(5)+B(0)+C(0)=1



Now Putting 3+2t=0









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