Step 1: Spot the hidden derivative on top.
Look at the numerator $\cos x - \sin x$. That is exactly what you get if you differentiate $\sin x + \cos x$, since $\frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x$. This is a strong hint to let that whole bracket be our new variable.
Step 2: Rewrite the bottom using the same bracket.
Square the bracket: $(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x\cos x = 1 + \sin 2x$. So $\sin 2x = (\sin x + \cos x)^2 - 1$.
Step 3: Replace $\sin 2x$ in the denominator.
The denominator $10 + \sin 2x$ becomes $10 + (\sin x + \cos x)^2 - 1 = 9 + (\sin x + \cos x)^2$. So the integral is now \[ I = \int \frac{\cos x - \sin x}{9 + (\sin x + \cos x)^2}\,dx \]
Step 4: Substitute $u = \sin x + \cos x$.
Then $du = (\cos x - \sin x)\,dx$, which is precisely the numerator times $dx$. The integral becomes the clean form \[ I = \int \frac{du}{9 + u^2} \]
Step 5: Use the standard arctan formula.
Recall $\int \frac{du}{a^2 + u^2} = \frac{1}{a}\tan^{-1}\!\left(\frac{u}{a}\right) + c$. Here $a^2 = 9$, so $a = 3$, giving $I = \frac{1}{3}\tan^{-1}\!\left(\frac{u}{3}\right) + c$.
Step 6: Put $u$ back.
Replace $u$ with $\sin x + \cos x$ to return to $x$. That is our answer.
\[ \boxed{\frac{1}{3}\tan^{-1}\left(\frac{\sin x + \cos x}{3}\right) + c} \]