Integrals of the Form $\int\dfrac{dx}{a\sin x+b\cos x+c}$

To evaluate $\int\dfrac{dx}{a\sin x+b\cos x+c}$, substitute $t=\tan\frac x2$. Then $\sec^2\frac x2\cdot\frac12\,dx=dt$, so $dx=\dfrac{2\,dt}{1+t^2}$ (using $\sec^2\frac x2=1+\tan^2\frac x2$); and, from the double-angle formulae applied to $\frac x2$, $\sin x=\dfrac{2t}{1+t^2}$ and $\cos x=\dfrac{1-t^2}{1+t^2}$. Substituting turns the whole integral into a rational function of $t$ with a quadratic denominator — the 3.2.4 type. For the closely related pattern $\int\dfrac{dx}{a\sin2x+b\cos2x+c}$, the substitution $t=\tan x$ is used instead, giving $dx=\dfrac{dt}{1+t^2}$, $\sin2x=\dfrac{2t}{1+t^2}$, $\cos2x=\dfrac{1-t^2}{1+t^2}$.

$\int\dfrac{dx}{5-4\cos x}$. With $t=\tan\frac x2$: denominator becomes $5-4\cdot\frac{1-t^2}{1+t^2}=\dfrac{9t^2+1}{1+t^2}$, so the integral is $\int\dfrac{2\,dt}{9t^2+1}=\frac29\int\dfrac{dt}{t^2+1/9}$, a Formula-1 shape: result $=\dfrac23\tan^{-1}\left(2\tan\dfrac x2\right)+c$.

$\int\dfrac{dx}{2-3\sin2x}$. With $t=\tan x$: denominator becomes $2-3\cdot\frac{2t}{1+t^2}=\dfrac{2t^2-6t+2}{1+t^2}$, giving $\int\dfrac{dt}{t^2-3t+1}$. Completing the square, $\left(t-\frac32\right)^2-\frac54$, a Formula-2 shape: result $=\dfrac{1}{2\sqrt5}\log\left|\dfrac{2\tan x-3-\sqrt5}{2\tan x-3+\sqrt5}\right|+c$.

$\int\dfrac{dx}{3-2\sin x+5\cos x}$. With $t=\tan\frac x2$, the denominator becomes $\dfrac{8-4t-2t^2}{1+t^2}$, giving $\int\dfrac{dt}{4-2t-t^2}$. Completing the square, $5-(t+1)^2$, a Formula-4-family shape ($\sqrt5$ vs $(t+1)$, log form since it's not under a root): result $=\dfrac{1}{2\sqrt5}\log\left|\dfrac{\sqrt5+1+\tan\frac x2}{\sqrt5-1-\tan\frac x2}\right|+c$. …