Question

Question #7f1cc

Answer 1

`int sqrt(x/(2-x))*dx=2arctansqrt(x/(2-x))-sqrt(2x-x^2)+C`

Explanation:

`int sqrt(x/(2-x))*dx`

After using `y=sqrt(x/(2-x))` substitution,

`y^2=x/(2-x)`

`x=y^2*(2-x)`

`x=2y^2-xy^2`

`(y^2+1)*x=2y^2`

`x=(2y^2)/(y^2+1)`

`dx=([4y*(y^2+1)-2y*2y^2]*dy)/(y^2+1)^2`

=`(4y*dy)/(y^2+1)^2`

After it, this integral became

`int y*(4y*dy)/(y^2+1)^2`

=`int (4y^2*dy)/(y^2+1)^2`

After using `y=tanz` and `dy=(secz)^2*dz` transormation, it became

`int [4(tanz)^2*(secz)^2*dz]/(secz)^4`

=`int 4(tanz)^2*(cosz)^2*dz`

=`int 4(sinz)^2*dz`

=`int (2-2cos2z)*dz`

=`2z-sin2z+C`

=`2z-(2tanz)/[(tanz)^2+1]+C`

After using `y=tanz` and `z=arctany` inverse transforms, I found

`2arctany-(2y)/(y^2+1)+C`

=`2arctansqrt(x/(2-x))-2sqrt(x/(2-x))/(x/(2-x)+1)+C`

=`2arctansqrt(x/(2-x))-sqrt[x*(2-x)]+C`

=`2arctansqrt(x/(2-x))-sqrt(2x-x^2)+C`