Evaluate

Asked by Rahulsinha1993 | 12th Sep, 2010, 12:27: PM

Expert Answer:

I=dxsinx +sin2x=dxsinx +2sinxcosx=dxsinx(1+2cosx)                         =dxsinx(1+2cosx).sinxsinx                          =sinx dxsin2x(1+2cosx)=sinx dx(1-cos2x)(1+2cosx)Let cosx =t, hence sinxdx = dtI = dt(1-t2)(1+2t)= dt(1+t)(1-t)(1+2t)Now use partial fractions,let 1(1+t)(1-t)(1+2t)=A1+t+B1-t+C1+2tHence A = -12, B= 16, C= 43I = -1/21+t+1/61-t+4/31+2t]dt = -12log(1+t)+16log(1-t)-1+43log(1+2t)2+c=-12log(1+sinx)-16log(1-sinx)+43log(1+2sinx)2+c">

Answered by  | 12th Sep, 2010, 02:53: PM

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