Given $v=\dfrac {x}{x^2+y^2} + \cos h \ y\cdot \cos x$

Steps:

1. For  given v we calculate $\psi1(x,y)=\dfrac{\partial v}{\partial y}=$vy

2. We then calculate  $\psi2(x,y)=\dfrac{\partial v}{\partial x}=$vx

3. By Milne thompson method we calculate $\psi1(z,0)$ and $ \psi2(z,0)$

4. Then f'(z)=$\psi1(z,0) + i \psi2(z,0)$

5. f(z)=$\int f'(z)\,\mathrm{d}z+c$

$\therefore$vy$=\psi1(x,y)$$=\dfrac {2xy}{(x^2+y^2)^2} + \sin h \ y\cdot \cos x$ $\therefore$vx$=\psi2(x,y)=\dfrac {-x^2+y^2}{(x^2+y^2)^2} - \cos h \ y\cdot \sin x$ By Milne Thompson method  $\psi1(z,0)=0 $  and $\psi2(z,0)=-sinz-\dfrac{1}{z^2}$......(putting x=z,y=0 in $\psi1$ and $\psi2$) Now f'(z)=$\psi1(z,0) + i \psi2(z,0)$ f(z)=$\int \psi1(z,0)\,\mathrm{d}z+i\int \psi2(z,0)\,\mathrm{d}z$ f(z) =-i $\int\bigg( \sin z+\dfrac{1}{z^2}\bigg)\,\mathrm{d}z$ f(z)=i $\bigg(\cos z +\dfrac{1}{z}\bigg)$+c