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Answer 1 · 2017-11-23T00:32:56

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

#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#