Step 1: Use symmetry of the interval:
Call the integral $I$. Change $x$ to $\frac{\pi}{2}-x$.
The denominator $1+\sin x\cos x$ stays the same, but $\sin^2 x$ turns into $\cos^2 x$.
So the integral with $\cos^2 x$ on top is also equal to $I$.
Step 2: Combine to remove the numerator:
Add the two equal integrals. The numerator is now $\sin^2 x+\cos^2 x=1$.
\[ 2I = \int_{0}^{\pi/2} \frac{dx}{1+\sin x\cos x} \]
Step 3: Write the denominator with a double angle:
Since $\sin x\cos x = \frac{\sin 2x}{2}$, the denominator is $\frac{2+\sin 2x}{2}$.
\[ 2I = \int_{0}^{\pi/2} \frac{2\,dx}{2+\sin 2x} \]
Put $u = 2x$, so $2\,dx = du$ and $u$ runs from $0$ to $\pi$.
\[ 2I = \int_{0}^{\pi} \frac{du}{2+\sin u} \]
Step 4: Use the half-angle substitution:
Let $t=\tan\frac{u}{2}$. Then $\sin u=\frac{2t}{1+t^2}$ and $du=\frac{2\,dt}{1+t^2}$.
The limits $u=0$ and $u=\pi$ become $t=0$ and $t\to\infty$.
\[ 2I = \int_{0}^{\infty} \frac{2\,dt}{2(1+t^2)+2t} = \int_{0}^{\infty} \frac{dt}{t^2+t+1} \]
Step 5: Complete the square and evaluate:
We have $t^2+t+1=\left(t+\frac12\right)^2+\frac34$.
The antiderivative is $\frac{2}{\sqrt3}\tan^{-1}\frac{2t+1}{\sqrt3}$.
At $\infty$ it gives $\frac{2}{\sqrt3}\cdot\frac{\pi}{2}$. At $0$ it gives $\frac{2}{\sqrt3}\cdot\frac{\pi}{6}$.
\[ 2I = \frac{2}{\sqrt3}\cdot\frac{\pi}{3} = \frac{2\pi}{3\sqrt3} \]
Final Answer:
Halving the result gives the required value.
\[ \boxed{I = \frac{\pi}{3\sqrt{3}}} \]