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a
-18log1-sinx1+sinx+142log1+2sinx1-2sinx+C
b
18log1+sinx1-sinx-142log1+2sinx1-2sinx+C
c
-18log1+sinx1-sinx+142log1+2sinx1-2sinx+C
d
None of these
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Correct option is C
let I=∫sin xsin 4xdx= ∫sin x2sin 2xcos2xdx=∫sin x4sin xcosx cos2xdx=14∫1cosx cos2xdx=14∫cosxcos2x cos2xdx=14∫cosx(1-sin2x)(1-2sin2x)dxputting sinx =t and cosxdx=dt, we getI=14∫dt(1-t2)(1-2t2)=14∫21-2t2-11-t2dt=-14∫dt(1-t2)+24∫dt1-(2t)2=-14x12log1+t1-t+12·122log1+2t1-2t+C =-18log1+sinx1-sinx+142log1+2sinx1-2sinx+CSimilar Questions
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