If x=x(y) is the solution of the differential equation, 2 · √(sin x+ tan^-1 y) · (d y)/(d x)=( cos x+1/(1+y^2) (d y)/(d x)) satisfying x(0)=0 and…

If $\displaystyle x=x(y)$ is the solution of the differential equation, $\displaystyle 2 \cdot \sqrt{\sin x+\tan ^{-1} y} \cdot \frac{d y}{d x}=\left(\cos x+\frac{1}{1+y^2} \frac{d y}{d x}\right)$ satisfying $\displaystyle x(0)=0$ and $\displaystyle x(1)=b$, then