Trigonome- what?

Note that $(1+\sin x)(1-\sin x)={\cos}^2 x.$

Then $\dfrac{\cos x}{1+\sin x}=\dfrac{2-2\sin x}{2\cos x}=\dfrac{2-2\sin x+\cos x}{1+\sin x+2\cos x},$ by using componendo .

Therefore, $\dfrac{2-2\sin x+\cos x}{1+\sin x+2\cos x}=\dfrac{\cos x}{1+\sin x}.$

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Let $1+\sin x=t,$ then $\cos x ~dx=dt.$

$\displaystyle \therefore~\left\lfloor 1000\int_{0}^{\frac{\pi}{2}} \dfrac{2-2\sin x+\cos x}{1+\sin x+2\cos x} dx \right\rfloor \\ \displaystyle =\left\lfloor 1000\int_{0}^{\frac{\pi}{2}} \dfrac{\cos x}{1+\sin x} dx \right\rfloor \\ \displaystyle =\left\lfloor 1000\int_{1}^{2} \dfrac{1}{t} dt \right\rfloor \\ =\left\lfloor 1000\left[\ln |t|\right]_{1}^{2} \right\rfloor \\ =\left\lfloor 1000\ln2 \right\rfloor \\ =\boxed{693}.$

We have $\begin{aligned} \int_0^{\frac12\pi} \frac{2\,dx}{1 + \sin x + 2\cos x} & = \; \int_0^{\frac12\pi} \frac{2\,dx}{1 + \cos x + 2\sin x} \; = \; \int_0^{\frac12\pi} \frac{dx}{\cos^2\tfrac12x + 2\sin\tfrac12x \cos\tfrac12x} \\ & = \; \int_0^{\frac12\pi} \frac{\sec^2\tfrac12x\,dx}{1 + 2\tan\tfrac12x} \; =\; \Big[\ln(1 +2\tan\tfrac12x)\Big]_0^{\frac12\pi} \; = \; \ln3 \end{aligned}$ while $\int_0^{\frac12\pi} \frac{-2\sin x + \cos x}{1 + \sin x + 2\cos x}\,dx \; = \; \Big[\ln(1 + \sin x + 2\cos x)\Big]_0^{\frac12\pi} \ = \; \ln\tfrac23$ Adding these together, the desired integral is $\ln2$ , making the answer $\lfloor 1000\ln2\rfloor = \boxed{693}$ .

H.M. and Mark Hennings had much better solutions than mine (and more creative as well), but I'll post mine anyway... I separated the fraction the same way Mark did. One of those integrals, namely, $\int_0^{\frac{\pi}{2}} \frac{-2\sin{x}+\cos{x}}{1+\sin{x}+2\cos{x}}dx$ was straightforward to compute, as the numerator was simply the derivative of the denominator. Its value is $\ln{2}-\ln{3}$ . I will be spending more time on $\int_0^{\frac{\pi}{2}} \frac{2}{1+\sin{x}+2\cos{x}} dx$ . I let $u=e^{ix}$ . This leads to $\int_1^i \frac{2}{1+\frac{1-\frac{1}{u}}{2i}+2\frac{u+\frac{1}{u}}{2}} \frac{du}{iu}$ . Skipping some steps, this can be manipulated into $\frac{4}{1+2i}\int_1^i \frac{du}{(u+\frac{2+i}{5})^2+(\frac{2}{5}(2+i))^2}=-2i\Big[\arctan(\frac{u(2-i)+1}{2})\Big]_1^i=-2i\Big(\arctan(1+i)-\arctan(\frac{3-i}{2})\Big)=-2i\Big(\frac{i}{2}\ln{\frac{i+1+i}{i-(1+i)}}-\frac{i}{2}\ln{\frac{i+\frac{3-i}{2}}{i-\frac{3-i}{2}}}\Big)=\ln{3}$ . So, the combined answer is $\ln{2}-\ln{3}+\ln{3}=\ln{2}$ . $\$ Here's another solution to $\int_0^{\pi/2} \frac{2}{1+\sin{x}+2\cos{x}}dx$ using the so-called Weierstrass substitution ( $t=\tan{\frac{x}{2}}$ ). Recall that with the Weierstrass substitution, $\cos{x}=\frac{1-t^2}{1+t^2}$ , $\sin{x}=\frac{2t}{1-t^2}$ , and $dx=\frac{2}{1+t^2}dt$ . Substituting these into the integral, I get $\int_0^1 \frac{2}{1+\frac{2t}{1+t^2}+2\frac{1-t^2}{1+t^2}}\frac{2}{1+t^2}dt=\int_0^1\frac{4}{-t^2+2t+3}dt$ $=\int_0^1\frac{1}{-t+3}+\frac{1}{t+1}dt=\Big[-\ln{(-t+3)}+\ln{t+1}\Big]_0^1=-\ln{2}+\ln{2}+\ln{3}-\ln{1}=\ln{3}$ .