1.

int(cos2x-cos2alpha)/(cosx-cosalpha)dx

Answer»

SOLUTION :`I=INT(cos2x-cos2alpha)/(cosx-cosalpha)DX`
=`int((2cos^2x-1)-(2cos^2alpha-1))/(cosx-cosalpha)dx`
=`2int(cosx+cosalpha)dx`
=`2sinx+2xcosalpha+c`


Discussion

No Comment Found

Related InterviewSolutions