1.

Evaluate `int(dx)/(cos xsqrt(cos2x))`.

Answer» `I=int(dx)/(cos xsqrt(cos2x))`
`=int(cosx dx)/(cos^(2) xsqrt(1-2sin^(2)x))`
`=int(cosx dx)/((1-sin^(2) x)sqrt(1-2sin^(2)x))`
`=int(dt)/((1-t^(2)) sqrt(1-2t^(2))) " "("Putting " sinx=t)`
Put `t=(1)/(y)`
` :. dt= -(1)/(y^(2))dy`
` :. I= -int(dy)/(y^(2)(1-(1)/(y^(2)))sqrt(1-(2)/(y^(2))))`
`= -int(ydy)/((y^(2)-1)sqrt(y^(2)-2))`
Now put `y^(2)-2=z^(2)`
` :. ydy=zdz`
` :.I= -int(dz)/(z^(2)+1)`
`= -tan^(-1)z+c`
`= -tan^(-1)sqrt(y^(2)-2)+c`
`= -"tan"^(-1)sqrt((1)/(t^(2))-2)+c`
`= -"tan"^(-1)sqrt((1)/(sin^(2)x)-2)+c`


Discussion

No Comment Found

Related InterviewSolutions