int (sinx)^2/(sinx+2cosx)*dx
After using u=tan(x/2), cosu=(1-u^2)/(u^2+1), sinu=(2u)/(u^2+1) and dx=(2du)/(u^2+1) transforms, this integral became
int [(2u)/(u^2+1)]^2/[2u/(u^2+1)+2*(1-u^2)/(u^2+1)]*(2du)/(u^2+1)
=int (-4u^2*du)/[(u^2+1)^2*(u^2-u-1)]
After decomposing integrand into basic fractions,
int (-4u^2*du)/[(u^2+1)^2*(u^2-u-1)]
=4/5*int (du)/(u^2+1)+4/5*int ((u-2)*du)/(u^2+1)^2-4/5*int (du)/(u^2-u-1)
=4/5*arctanu+2/5*int (2u*du)/(u^2+1)^2-8/5*int (du)/(u^2+1)^2-16/5*int (du)/(4u^2-4u-4)
=4/5*arctanu-2/5*(u^2+1)^(-1)+C1-8/5*int (du)/(u^2+1)^2-8/5*int (2du)/[(2u-1)^2-5]
I=int (du)/(u^2+1)^2
After taking u=tany and du=(secy)^2*dz transormation, I became
I=int ((secy)^2*dy)/(secy)^4
=int (cosy)^2*dy
=int 1/2*(1+cos2y)*dy
=1/2*y+1/4*sin2y
After using u=tany, y=arctanu and sin2y=2u/(u^2+1) inverse transforms, I found
I=int (du)/(u^2+1)^2=1/2*arctanu+1/2*u/(u^2+1)
J=int (2du)/[(2u-1)^2-5]
After taking 2u-1=z and 2du=dz transforms, J became
J=int dz/(z^2-5)
=int dz/[(z+sqrt5)*(z-sqrt5)]
=sqrt5/10*int dz/(z-sqrt5)-sqrt5/10*int dz/(z+sqrt5)
=sqrt5/10*Ln(z-sqrt5)-sqrt5/10*Ln(z+sqrt5)
=sqrt5/10*Ln[(z-sqrt5)/(z+sqrt5)]
==sqrt5/10*Ln[(2u-1-sqrt5)/(2u-1+sqrt5)]
Hence,
int (-4u^2*du)/[(u^2+1)^2*(u^2-u-1)]
=4/5*arctanu-2/5*(u^2+1)^(-1)+C1-8/5*[1/2*arctanu+1/2*u/(u^2+1)]
-8/5*sqrt5/10*Ln[(2u-1-sqrt5)/(2u-1+sqrt5)]
=4/5*artanu-2/5*(u^2+1)^(-1)-4/5arctanu-4/5*u/(u^2+1)-(4sqrt5)/25*Ln[(2u-1-sqrt5)/(2u-1+sqrt5)]+C1
=-2/5*(u^2+1)^(-1)-4/5*u/(u^2+1)-(4sqrt5)/25*Ln[(2u-1-sqrt5)/(2u-1+sqrt5)]+C1
Thus,
int (sinx)^2/(sinx+2cosx)*dx
=-2/5*[(tan(x/2)^2+1]^(-1)-4/5*tan(x/2)/[(tan(x/2))^2+1)]-(4sqrt5)/25*Ln[(2tan(x/2)-1-sqrt5)/(2tan(x/2)-1+sqrt5)]+C1
=-2/5*(cos(x/2))^2-2/5*sinx-(4sqrt5)/25*Ln[(2tan(x/2)-1-sqrt5)/(2tan(x/2)-1+sqrt5)]+C1
=-1/5*cosx-2/5*sinx-(4sqrt5)/25*Ln[(2tan(x/2)-1-sqrt5)/(2tan(x/2)-1+sqrt5)]+C
Note: C=C1-1/5
