Ytukyg

How do you integrate $int frac{1}{a + cos x} dx$? Is it solvable by elementary methods? I was trying to do it while incorrectly solving a homework problem but I couldn't find the answer.

Thanks!

Let $ y = frac{x}{2}$.

$$frac{1}{a + cos 2y} = frac{1}{a -1 + 2cos ^2 y} = frac{sec^2 y}{(a-1)sec^2 y + 2} = frac{sec^2 y}{a + 1 + (a-1)tan^2 y} $$

Thus

$$int frac{1}{a + cos x} text{d}x = int frac{2}{a + cos 2y} text{d}y $$

$$ = int frac{ 2sec^2 y}{ a + 1 + (a-1)tan^2 y} text{d} y$$

Now make the subsitution $t = tan y$.

I remember having used the same trick before: Summing the series $ frac{1}{2n+1} + frac{1}{2} cdot frac{1}{2n+3} + cdots text{ad inf}$

Generalization:

Let's consider $cos x = frac{1-t^2}{1+t^2}; t = tanfrac{x}{2}; dx=frac{2}{1+t^2} dt.$ Then we get that our integral becomes:

$$J = int frac{2dt}{(a+b)+(a-b) t^2}$$

I. For the case $a>b$, consider $a+b=u^2$ and $a-b=v^2$, and obtain that:
$$J = 2int frac{dt}{u^2+v^2 t^2}=frac{2}{uv} arctanfrac{vt}{u} +C.$$
Turning back to our notation we get:
$$I=frac{2}{sqrt{a^2-b^2}} arctanleft(sqrt{frac{a-b}{a+b}} tanfrac{x}{2} right) + C.$$

II. For the case $a<b$, consider $a+b=u^2$ and $a-b=-v^2$, and obtain that:
$$J = 2int frac{dt}{u^2-v^2 t^2}=frac{1}{uv}lnfrac{u+vt}{u-vt} +C.$$

Turning back again to out initial notation and have that:

$$I=frac{2}{sqrt{b^2-a^2}} lnfrac{b+a cos x + sqrt{b^2-a^2} sin x}{a+b cos x} + C.$$
Also, note that $x$ must be different from ${+}/{-}arccos(-frac{a}{b})+2kpi$ if $|frac{a}{b}|leq1$.

Q.E.D.

This doesn't help you to evaluate the indefinite integral, but I though I would add that the definite integral $int_0^{2 pi
} frac{1}{a + cos x} dx$ can also be evaluated using methods from complex analysis. Let us make the simplifying assumption that $a > 1$ to avoid a blow up in the integral. Other values of $a$ (including complex ones!) can be made to work too.

We have
begin{align*}
int_0^{2 pi
} frac{dx}{a + cos x} &= int_0^{2 pi} frac{dx}{a + frac{e^{ix} + e^{-ix}}{2}} \
&= 2int_0^{2 pi} frac{e^{ix} dx}{2ae^{ix} + e^{2ix} + 1} && text{Let } z=e^{ix}, text{ so } dz = ie^{ix} dx. \
&= frac{2}{i} int_{|z|=1} frac{dz}{z^2 + 2az + 1} \
&= frac{2}{i} int_{|z|=1} frac{dz}{(z-z_1)(z-z_2)}
end{align*}
where the circle $|z|=1$ is parametrized counterclockwise and
begin{align*} z_1 = -a - sqrt{a^2-1} && z_2 = -a + sqrt{a^2-1}.
end{align*}
Clearly $z_1 < -a < -1$, so $z_1$ is outside of the unit circle. As for $z_2$, it is convenient to notice that
begin{align*}
z_1 z_2 = 1.
end{align*}
Thus, since $z_1$ is outside the unit circle, its inverse $z_2$ is inside the unit circle. So, applying the residue theorem, we have

begin{align*}
frac{2}{i} int_{|z|=1} frac{dz}{(z-z_1)(z-z_2)}
&= frac{2}{i} 2 pi i mathrm{Res}left( frac{1}{(z-z_1)(z-z_2)}, z_2right) \
&= 4 pi frac{1}{z_2 - z_1} \
&= 4 pi frac{1}{2 sqrt{a^2 -1}} \
&= frac{2 pi}{sqrt{a^2-1}}.
end{align*}

One can check this result against the one obtained from Arybatha's antiderivative, although a bit of care needs to be taken as an improper integral crops up while making the various substitutions.
begin{align*}
int_0^{2 pi
} frac{dx}{a + cos x}
&= int_0^pi frac{2 dy}{a+1 + (a-1) tan^2(y)} && y= frac{x}{2}\
&= int_0^infty frac{ 2 dt}{ a + 1 + (a-1)t^2} +int_{-infty}^0 frac{ 2 dt}{ a + 1 + (a-1)t^2} && t = tan(y) \
&= frac{2}{sqrt{a^2-1}} arctanleft( sqrt{ frac{a-1}{a+1}} t right) big|_0^infty + frac{2}{sqrt{a^2-1}} arctanleft( sqrt{ frac{a-1}{a+1}} t right) big|_{-infty}^0 \
&= frac{2 pi}{sqrt{a^2-1}}
end{align*}

Expanding André's comment

Say we have an integral of the form

$$int R(sin x,cos x) dx$$

Then the substitution

$$t= tanfrac x 2 $$

will change the integral into a rational function of

$$sin x = frac{2t}{1+t^2}$$

$$cos x = frac{1-t^2}{1+t^2}$$

and of course

$$dx = frac{2 dt}{1+t^2}$$

Would you like to try solve it that way or want a full solution?

Also a generalised solution, borrowing from and expanding upon user1357113's answer,

I. For the case $|a| > |b|$, note that the substitution $t=tan left( frac{x}{2} right)$ is not injective. So to retrieve a continuous antiderivative, we have as a complete answer

$$
int frac{1}{a+bcos(x)} {rm d}x
= frac{2}{sqrt{a^2-b^2}} arctan left( sqrt{frac{a-b}{a+b}}tan left( frac{x}{2} right) right) + frac{2pi}{sqrt{a^2-b^2}} leftlfloor frac{x+pi}{2pi} rightrfloor +C
$$

However, we can simplify the ugly floor function in terms of arctangents and tangents. First, we have

$$
pi leftlfloor frac{x+pi}{2pi} rightrfloor
= frac{x}{2} - arctan left( tan left( frac{x}{2} right) right)
$$

So, implementing this and then asking What is $arctan(x) + arctan(y)$? We get

$$
begin{align}
int frac{1}{a+bcos(x)} {rm d}x
& = frac{2}{sqrt{a^2-b^2}} arctan left( sqrt{frac{a-b}{a+b}}tan left( frac{x}{2} right) right) + frac{2}{sqrt{a^2-b^2}} left( frac{x}{2} - arctan left( tan left( frac{x}{2} right) right) right) +C \
& = frac{1}{sqrt{a^2-b^2}} left( x + 2arctan left( sqrt{frac{a-b}{a+b}}tan left( frac{x}{2} right) right) - 2arctan left( tan left( frac{x}{2} right) right) right) + C \
& = frac{1}{sqrt{a^2-b^2}} left( x + 2arctan left( frac{ sqrt{frac{a-b}{a+b}}tan left(frac{x}{2}right) - tan left(frac{x}{2}right) }{1 + sqrt{frac{a-b}{a+b}} tan^2 left(frac{x}{2}right) } right) right) + C \
end{align}
$$

Which ultimately simplifies to a satisfying compact

$$ int frac{1}{a+bcos(x)} {rm d}x
= frac{1}{sqrt{a^2-b^2}} left( x - 2arctan left( frac{ (sqrt{a+b}-sqrt{a-b})tan left(frac{x}{2}right) }{sqrt{a+b} + sqrt{a-b} tan^2 left(frac{x}{2}right) } right) right) + C
$$

II. And for the case $|a|<|b|$, we will have, from user1357113's partial fraction decomposition,

$$ int frac{1}{a+bcos(x)} {rm d}x
= frac{1}{sqrt{b^2-a^2}}lnleft(left|frac{b+acosleft(xright)+sqrt{b^2-a^2}sinleft(xright)}{a+bcosleft(xright)}right|right) +C
$$

III a. And for the case $a = b$, we will have
$$
begin{align}
int frac{1}{a+bcos(x)} {rm d}x
& = frac1a int frac{1}{1+cos(x)} {rm d}x \
& = frac1a int frac{1}{1+2cos^2(frac{x}{2})-1}{rm d}x \
& = frac1{2a} int sec^2 left(frac{x}{2}right) {rm d}x \
& = frac1a tanleft(frac{x}{2}right) + C
end{align}
$$

III b. Finally, for the case $a = -b$, we will have
$$
begin{align}
int frac{1}{a+bcos(x)} {rm d}x
& = frac1a int frac{1}{1-cos(x)} {rm d}x \
& = frac1a int frac{1}{1 - 1 + 2sin^2(frac{x}{2})}{rm d}x \
& = frac1{2a} int csc^2 left(frac{x}{2}right) {rm d}x \
& = frac1b cotleft(frac{x}{2}right) + C
end{align}
$$