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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` |
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