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