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Question

The value of the integral  cos3x+cos5xsin2x+sin4x dx is 
  1. sin x2(sin x)16tan1(sin x)+c
  2.  sin x6tan1(sin x)+c
  3. sin x2(sin x)1+c
  4. sin x2(sin x)1+5tan1(sin x)+c


Solution

The correct option is A sin x2(sin x)16tan1(sin x)+c
Let I = cos3x+cos5xsin2x+sin4xdx=(cos2x+cos4x)cosxsin2x(1+sin2x)dx=[1sin2x+(1sin2x)2]cosxsin2x(1+sin2x)dx=(23sin2x+sin4)cosxsin2x(1+sin2x)dxput sinx=tcosxdx=dtI=23t2+t4t4+t2dtI=(1+2t26t2+1)dt=t2t6tan1(t)+c=sinx2(sin x)16tan1(sinx)+c

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