Q.Integrate: $\int \frac{1}{4-5\cos x}\,dx$

Use the Weierstrass substitution $t=\tan(x/2)$.

Let $t=\tan\dfrac x2$, so $\cos x=\dfrac{1-t^2}{1+t^2}$, $\sin x=\dfrac{2t}{1+t^2}$, $dx=\dfrac{2\,dt}{1+t^2}$.

$2+\cos x-\sin x = 2+\dfrac{1-t^2}{1+t^2}-\dfrac{2t}{1+t^2} = \dfrac{2(1+t^2)+1-t^2-2t}{1+t^2} = \dfrac{t^2-2t+3}{1+t^2}$

$\int\dfrac{dx}{2+\cos x-\sin x} = \int\dfrac{\frac{2\,dt}{1+t^2}}{\frac{t^2-2t+3}{1+t^2}} = \int\dfrac{2\,dt}{t^2-2t+3} = \int\dfrac{2\,dt}{(t-1)^2+2}$

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Use the Weierstrass substitution $t=\tan(x/2)$ to convert the trigonometric integrand into a rational function of $t$.

Let $t=\tan(x/2)$, so $\sin x=\dfrac{2t}{1+t^2}$, $\cos x=\dfrac{1-t^2}{1+t^2}$, $dx=\dfrac{2\,dt}{1+t^2}$.

The denominator becomes:

$3+2\sin x+\cos x = 3+\frac{4t}{1+t^2}+\frac{1-t^2}{1+t^2} = \frac{3(1+t^2)+4t+1-t^2}{1+t^2} = \frac{2t^2+4t+4}{1+t^2}$

So the integral becomes:

$\int \frac{1+t^2}{2t^2+4t+4}\cdot\frac{2\,dt}{1+t^2} = \int\frac{2\,dt}{2t^2+4t+4} = \int\frac{dt}{t^2+2t+2}$

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