Better way to evaluate $\int \frac{dx}{\left (a +b\cos x \right)^2}$

Question

$$\int \frac{dx}{\left (a +b\cos x \right)^2}$$ $$u=\frac{b +a \cos x}{a +b\cos x }$$ $$du=\frac{\sin x\left(b^2 -a^2\right)}{ \left (a +b.\cos x \right)^2}$$ $$\frac{du}{\sin x\left(b^2 -a^2\right)}=\frac{dx}{\left (a +b\cos x \right)^2}$$ $$\cos x=\frac{au -b}{a - bu} $$ $$\sin x=\sqrt{1-\left(\frac{au -b}{a - bu}\right)^2}$$ It is becoming messy with this .I also tried it using half angle formula but didn't find it good.

Answer 1

HINT Put $$A=\frac{\ sin x}{a+b\cos x} $$now differentiate both sides and after simplification integrate it up to get the desired result.

Answer 2

I think the tangent half-angle substitution should work. Here's what I have:

\begin{align*} \int \frac{\mathrm{d}x}{(a+b\cos{x})^2} &= \int \frac{2\,\mathrm{d}t}{(1+t^2)(a+b\frac{1-t^2}{1+t^2})^2}, \qquad t = \tan{(x/2)} \\ &= 2 \int \frac{(1+t^2)dt}{a^2(1+t^2)^2+2ab(1-t^2)(1+t^2)+b^2(1-t^2)^2}\\ &= 2\int \frac{(1+t^2)\mathrm{d}t}{(a^2-2ab+b^2)t^4 + 2(a^2-b^2)t^2+a^2+2ab+b^2}\\ &=2\int \frac{(1+t^2)\mathrm{d}t}{((a-b)t^2 + (a+b))^2} \\ &=2\int \frac{(1+\frac{a+b}{a-b}\tan^2{\theta})\sqrt{\frac{a+b}{a-b}}\sec^2{\theta} \, \mathrm{d}\theta}{(a+b)^2\sec^4{\theta}} \qquad t=\sqrt{\frac{a+b}{a-b}}\tan{\theta} \\ &= 2\sqrt{\frac{a+b}{a-b}}\frac{1}{(a+b)^2}\int(\cos^2{\theta}+\frac{a+b}{a-b}\sin^2{\theta})\mathrm{d}\theta \\ &= \sqrt{\frac{a+b}{a-b}}\frac{1}{(a+b)^2}\left[\frac{2a}{a-b}\theta - \frac{2b}{a-b}\sin{\theta}\cos{\theta}\right] \end{align*}

Answer 3

Hint:

Rationalize with $z=e^{ix}$ (and $dz=iz\,dx$),

$$\int \frac{dx}{\left (r+\cos x \right)^2} =\int\frac{dz}{iz\left(r+\dfrac{z+z^{-1}}2\right)^2} =\int\frac{4z\,dz}{i\left(z^2+2rz+1\right)^2} =\int\frac{4(z+r-r)\,dz}{i\left((z+r)^2+1-r^2\right)^2}. $$

This yields a term $\log({z^2+2rz+1})=\log(r+\cos x)$ and, depending on the sign of $1-r^2$, an $\arctan$ or $\text{artanh}$.