Differentiating both sides of the given equation and noting the fundamental theorem of calculus yield

$$f'(x)\cos{x}-f(x)\sin{x}+2f(x)\sin{x}=1 \implies  f'(x)+f(x)\tan{x}=\frac{1}{\cos{x}}$$

This is a standard equation which can be solved using the integrating factor method. Here, the integrating factor equals

$$u(x)=e^{\int\tan{x}d{x}}=e^{\int\frac{\sin{x}}{\cos{x}}d{x}}=e^{-\int\frac{1}{\cos{x}}}d{\cos{x}}=e^{-\ln{\cos{x}}}=\frac{1}{\cos{x}}$$

Setting this to the solution formula  gives

$$f(x)=\cos{x}\left(\int\frac{1}{\cos^2{x}}d{x}+C\right)=\cos{x}\left(\tan{x}+C\right)=\sin{x} + C\cos{x}$$

where $C$ is a constant. Now, setting $x=0$ to the originally given equation leads to $f(0) = 1$. Therefore $C$ must equal $1$. Hence, the final answer is $$f(x) =\boxed{ \sin{x} + \cos{x}}$$