Any different way to solve an integral involving a trigonometric ratio?

Question

Recently I posted a question regarding computation of an interesting definite integral that involved trigonometric functions. Ever since, the topic of trigonometric integrals has quite attracted me, and I decided to sift through some calculus books that I have in search of interesting and challenging problems to solve.

Yesterday I stumbled upon one that required me to compute the following integral: $$ \mathcal{I}(0, 2\pi) = \int\limits_0^{2\pi}\frac{\sin(x)+1}{\cos(x) +2}dx $$ It was quite challenging for me to deal with this integral, so I decided to refer to WolframAlpha, which suggested I use the substitution $u=\tan(x/2)$ to attempt the integral. And it seems like the final answer should be $\mathcal{I}(0, 2\pi)=\dfrac{2\pi\sqrt{3}}{3}$.

Surely, since the function under the integral is defined everywhere in $\mathbb{R}$ (with period $\mathrm{lcm}(2\pi,2\pi)=2\pi$) the above substitution proposal seems valid. Yet I wonder whether there is some other “smarter” analytical approach to deal with this integral apart from tangential substitution.

Any ideas or suggestions will be greatly appreciated.

Answer 1

Decompose the integrand as follows \begin{align} & \int_0^{2\pi}\frac{\sin x+1}{\cos x +2}\ dx\\ =& \int_ 0^{2\pi}\left( \frac{\sin x}{\cos x +2}- \frac{\cos x+2-\sqrt3}{\sqrt3(\cos x +2)}+\frac1{\sqrt3}\right)dx\\ =& \bigg[\ln(\cos x+2)-2\tan^{-1}\frac{\sin x}{\cos x+2+\sqrt3}+\frac x{\sqrt3}\bigg]_0^{2\pi}=\frac{2\pi}{\sqrt3} \end{align} where the first two terms vanish due to periodicity.

Answer 2

If the contest-math tag is actually relevant, the contour integration approach probably is not your best choice. But your integral makes for a rather simple example:

$$\begin{align*} & \int_0^{2\pi} \frac{\sin x+1}{\cos x+2} \, dx \\ &= \int_0^{2\pi} \frac{\frac{e^{ix}-e^{-ix}}{2i} + 1}{\frac{e^{ix}+e^{-ix}}2+2} \, dx \\ &= \oint\limits_{\lvert z\rvert=1} \frac{\frac{z-\frac1z}{2i}+1}{\frac{z+\frac1z}2+2} \, \frac{dz}{iz} \\ &= - \oint\limits_{\lvert z\rvert=1} \underbrace{\frac{z^2+2iz-1}{z \left(z+2+\sqrt3\right) \left(z+2-\sqrt3\right)}}_{=:f(z)} \, dz \\ &= -i2\pi \sum_{z_0\in\left\{0,-2+\sqrt3\right\}} \underset{z=z_0}{\operatorname{Res}} f(z) \\ &= -i2\pi \left(-1 + 1+\frac i{\sqrt3}\right) = \frac{2\pi}{\sqrt3} \end{align*}$$

Answer 3

This is probably not a very intuitive solution, but if one happens to know this Fourier Series, then

$$ \begin{align} I &:= \int_{0}^{2\pi}\frac{\sin x+1}{\cos x+2}dx \\ &= \int_{0}^{2\pi}\frac{\sin x}{\cos x+2}dx+\int_{0}^{2\pi}\frac{1}{\cos x+2}dx \\ &= -\int_{1}^{1}\frac{1}{u}du+\int_{0}^{2\pi}\left(\frac{1}{\sqrt{2^{2}-1^{2}}}+\frac{2}{\sqrt{2^{2}-1^{2}}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{2^{2}-1^{2}}-2}{1}\right)^{n}\cos\left(nx\right)\right)dx \\ \require{cancel} &= 0+\frac{2\pi}{\sqrt{3}}+\frac{2}{\sqrt{3}}\sum_{n=1}^{\infty}\left(\sqrt{3}-2\right)^{n}\cancelto{0}{\int_{0}^{2\pi}\cos\left(nx\right)dx} \\ &= \frac{2\pi}{\sqrt{3}} \,. \\ \end{align} $$

Answer 4

Weierstrass Substitution $$ \begin{align} \int_0^{2\pi}\frac{\sin(x)+1}{\cos(x)+2}\,\mathrm{d}x &=\int_0^{2\pi}\frac{-\sin(x)+1}{\cos(x)+2}\,\mathrm{d}x\tag{1a}\\ &=\int_0^{2\pi}\frac1{\cos(x)+2}\,\mathrm{d}x\tag{1b}\\ &=\int_{-\infty}^\infty\frac{1}{\frac{1-z^2}{1+z^2}+2}\frac{2\,\mathrm{d}z}{1+z^2}\tag{1c}\\ &=\int_{-\infty}^\infty\frac2{3+z^2}\,\mathrm{d}z\tag{1d}\\ &=\frac2{\sqrt3}\int_{-\infty}^\infty\frac1{1+z^2}\,\mathrm{d}z\tag{1e}\\ &=\frac{2\pi}{\sqrt3}\tag{1f} \end{align} $$ Explanation:
$\text{(1a):}$ substitute $x\mapsto2\pi-x$
$\text{(1b):}$ average both sides of $\text{(1a)}$
$\text{(1c):}$ Weierstrass Substitution
$\text{(1d):}$ simplify
$\text{(1e):}$ substitute $z\mapsto\sqrt3z$
$\text{(1f):}$ apply arctangent integral

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Contour Integration $$ \begin{align} \int_0^{2\pi}\frac{\sin(x)+1}{\cos(x)+2}\,\mathrm{d}x &=\int_0^{2\pi}\frac{-\sin(x)+1}{\cos(x)+2}\,\mathrm{d}x\tag{2a}\\ &=\int_0^{2\pi}\frac1{\cos(x)+2}\,\mathrm{d}x\tag{2b}\\ &=\int_{|z|=1}\frac2{z+\frac1z+4}\frac{\mathrm{d}z}{iz}\tag{2c}\\ &=\frac2i\int_{|z|=1}\frac1{z^2+4z+1}\,\mathrm{d}z\tag{2d}\\ &=4\pi\operatorname*{Res}_{z=-2+\sqrt3}\left(\frac1{z^2+4z+1}\right)\tag{2e}\\ &=\left.4\pi\frac1{2z+4}\right|_{z=-2+\sqrt3}\tag{2f}\\ &=\frac{2\pi}{\sqrt3}\tag{2g} \end{align} $$ Explanation:
$\text{(2a):}$ substitute $x\mapsto2\pi-x$
$\text{(2b):}$ average both sides of $\text{(2a)}$
$\text{(2c):}$ along the contour, $z=e^{ix}$,
$\phantom{\text{(2c):}}$ $\cos(x)=\frac12\left(z+\frac1z\right)$
$\phantom{\text{(2c):}}$ and $\mathrm{d}x=\frac{\mathrm{d}z}{iz}$
$\text{(2d):}$ simplify
$\text{(2e):}$ apply the Residue Theorem
$\phantom{\text{(2e):}}$ the singularity inside the unit circle
$\phantom{\text{(2e):}}$ is at $z=-2+\sqrt3$
$\text{(2f):}$ if $\frac1{f(z_0)}$ is a simple pole,
$\phantom{\text{(2f):}}$ then the residue is $\frac1{f'(z_0)}$
$\text{(2f):}$ evaluate

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Double Angle Formula $$ \begin{align} \int_0^{2\pi}\frac{\sin(x)+1}{\cos(x)+2}\,\mathrm{d}x &=\int_0^{2\pi}\frac{-\sin(x)+1}{\cos(x)+2}\,\mathrm{d}x\tag{3a}\\ &=\int_{-\pi}^\pi\frac1{\cos(x)+2}\,\mathrm{d}x\tag{3b}\\ &=\int_{-\pi}^\pi\frac1{2\cos^2(x/2)+1}\,\mathrm{d}x\tag{3c}\\ &=\int_{-\pi}^\pi\frac2{2+\sec^2(x/2)}\,\mathrm{d}\tan(x/2)\tag{3d}\\ &=\int_{-\infty}^\infty\frac2{3+u^2}\,\mathrm{d}u\tag{3e}\\ &=\frac2{\sqrt3}\int_{-\infty}^\infty\frac1{1+u^2}\,\mathrm{d}u\tag{3f}\\ &=\frac{2\pi}{\sqrt3}\tag{3g} \end{align} $$ Explanation:
$\text{(3a):}$ substitute $x\mapsto2\pi-x$
$\text{(3b):}$ average both sides of $\text{(3a)}$
$\phantom{\text{(3b):}}$ use the $2\pi$ periodicity of $\cos(x)$
$\text{(3c):}$ $\cos(x)=2\cos^2(x/2)-1$
$\text{(3d):}$ $\mathrm{d}x=2\cos^2(x/2)\,\mathrm{d}\tan(x/2)$
$\text{(3e):}$ $u=\tan(x/2)$
$\text{(3f):}$ substitute $u\mapsto\sqrt3u$
$\text{(3g):}$ evaluate

Answer 5

In general, it is often very helpful to use the Weierstrass substitution $x=\tan(u/2)$, as it can turn trig integrals into rational integrals. However, if you want to avoid it, this method works(although I would not be surprised if there is a shorter method, this one is pretty long for a contest problem I think).

To start, we use King's rule, which states that $\int_{a}^{b} f(x) \text{ dx} = \int_{a}^{b} f(a+b-x) \text{ dx}$

This gives us:

\begin{align} & 2I(0, 2\pi) = \int_{0}^{2 \pi} \frac{\sin(x)+1}{\cos(x)+2} \text{ dx} + \int_{0}^{2 \pi} \frac{\sin(2 \pi + 0 - x)+1}{\cos(2 \pi + 0 - x)+2} \text{ dx} =\\ & \int_{0}^{2 \pi} \frac{\sin(x)+1}{\cos(x)+2} + \frac{\sin(2 \pi + 0 - x)+1}{\cos(2 \pi + 0 - x)+2} \text{ dx} =\\ & \int_{0}^{2 \pi} \frac{\sin(x)+1}{\cos(x)+2} + \frac{-\sin(x)+1}{\cos(x)+2} \text{ dx} =\\ & \int_{0}^{2 \pi} \frac{\sin(x)-\sin(x)+2}{\cos(x)+2} \text{ dx}=\\ & \int_{0}^{2 \pi} \frac{2}{\cos(x)+2} \text{ dx}\\ \end{align}

So, we have that $I(0, 2 \pi) = \int_{0}^{2 \pi} \frac{1}{\cos(x)+2} \text{ dx}$. Now, it is easy to see that $\int_{0}^{\pi} \frac{1}{\cos(x)+2} \text{ dx} = \int_{\pi}^{2 \pi} \frac{1}{\cos(x)+2} \text{ dx}$, so we have that $1/2 \cdot I(0, 2 \pi) = \int_{0}^{\pi} \frac{1}{\cos(x)+2} \text{ dx}$.

We can utilize King's rule again on this integral, which gives us

$\int_{0}^{\pi} \frac{1}{\cos(x)+2} \text{ dx} = \int_{0}^{\pi} \frac{1}{\cos(\pi + 0 -x)+2} \text{ dx}$

and then

\begin{align} & I(0, 2 \pi) = \int_{0}^{\pi} \frac{1}{\cos(x)+2} + \frac{1}{\cos(\pi + 0 -x)+2} \text{ dx}=\\ & \int_{0}^{\pi} \frac{1}{\cos(x)+2} + \frac{1}{-\cos(x)+2} \text{ dx}=\\ & \int_{0}^{\pi} \frac{4}{4-\cos^2(x)} \text{ dx}=\\ \end{align}

Now, there is a very neat trick here that allows one to evaluate this integral. We start by multiplying numerator and denominator by $\sec^2(x)$ and using the identity $\sec^2(x)-1=\tan^2(x)$:

\begin{align} & I(0, 2 \pi) = \int_{0}^{\pi} \frac{4}{4-\cos^2(x)} \text{ dx}=\\ & 4 \int_{0}^{\pi} \frac{\sec^2(x)}{4 \sec^2(x)-1} \text{ dx}=\\ & 4 \int_{0}^{\pi} \frac{\sec^2(x)}{4(\tan^2(x)+1)-1} \text{ dx}=\\ & 4 \int_{0}^{\pi} \frac{\sec^2(x)}{4 \tan^2(x)+3} \text{ dx}\\ \end{align}

Here, we must be careful. The denominator is not defined at $x=\frac{\pi}{2}$, so we must integrate around this as follows:

\begin{align} & 1/4 \cdot I(0, 2\pi) = \int_{0}^{\pi/2} \frac{\sec^2(x)}{4 \tan^2(x)+3} \text{ dx} + \int_{\pi/2}^{\pi} \frac{\sec^2(x)}{4 \tan^2(x)+3} \text{ dx} \end{align}

and now the substitution $u=\tan(x)$, $du=\sec^2(x) dx$ is valid:

\begin{align} & 1/4 \cdot I(0, 2 \pi) = \int_{0}^{\infty} \frac{1}{4 u^2 + 3} \text{ du} + \int_{- \infty}^{0} \frac{1}{4 u^2 + 3} \text{ du}=\\ & \int_{-\infty}^{\infty} \frac{1}{4 u^2 + 3} \text{ du}=\\ & \frac{1}{3} \int_{-\infty}^{\infty} \frac{1}{(\frac{2u}{\sqrt{3}})^2 + 1} \text{ du} \end{align}

and a substitution $v=\frac{2u}{\sqrt{3}}$, $dv=\frac{2}{\sqrt{3}} du$ gives

\begin{align} & 1/4 \cdot I(0, 2 \pi) = \frac{1}{2 \sqrt{3}}\int_{-\infty}^{\infty} \frac{1}{v^2+1} \text{ dv}=\\ & \frac{1}{2 \sqrt{3}} (\arctan(\infty) - \arctan(- \infty))=\\ & \frac{1}{2 \sqrt{3}} \pi \end{align}

So, we have that $\boxed{I(0, 2 \pi) = \frac{2 \pi}{\sqrt{3}} = \frac{2 \pi \sqrt{3}}{3}}$

Answer 6

Both $\sin$ and $\cos$ have period $2\pi$ so $\int_{\pi}^{2\pi}\frac{\sin(x)+1}{\cos(x) +2}dx = \int_{-\pi}^{0}\frac{\sin(x)+1}{\cos(x) +2}dx$ and your integral can be written more symmetrically as

$$\int_{-\pi}^{\pi}\frac{\sin(x)}{\cos(x) +2} + \frac{1}{\cos(x) +2}dx$$

But the first term is an odd function, so its integral from $-\pi$ to $\pi$ vanishes. The second term is an even function, with the same integral from $-\pi$ to $0$ as it has from $0$ to $\pi$. So what you want to find is

$$\int_{0}^{\pi}\frac{2}{\cos(x) +2}dx$$

You've already seen some ways to solve this. Here's another, for completeness. Note that for $0 \le x \le \pi$ we have $\sin x \ge 0$ so $\sin x = \sqrt{1 - \cos^2 x}$ (i.e. we only need to take the positive root). Making the substitution $u = \cos x$, $\frac{du}{dx} = -\sin x = - \sqrt{1-u^2}$, we get

$$\int_{u=1}^{u=-1} \frac{2}{u+2} \cdot \frac{-1}{\sqrt{1-u^2}} du = \int_{-1}^1 \frac{2}{(u+2)\sqrt{1-u^2}}du$$

An integral of the form $\int R(u, \sqrt{au^2 + bu + c}) du$, where $R$ is a rational function, is amenable to one of the Euler substitutions. In our case, $au^2 + bu + c = 1 - u^2$. Euler's first substitution works if $a > 0$, but sadly for us $a = -1$. Euler's second substitution works if $c > 0$, which is fine since we have $c = 1$. Alternatively, Euler's third substitution works if $au^2 + bu + c$ has real roots $\alpha$ and $\beta$; we can do this with $\alpha = 1$ and $\beta = -1$.

Euler's second substitution: putting $\sqrt{au^2 + bu + c} = ut \pm \sqrt{c}$ and taking the positive sign gives, in our case, $\sqrt{1-u^2} = ut + 1$, or $t = \frac{\sqrt{1-u^2} - 1}{u}$. This rearranges to $$u = \frac{\pm 2t \sqrt{c} - b}{a - t^2} = \frac{2t}{-1-t^2} = \frac{-2t}{1+t^2}$$

and the derivative is

$$\frac{du}{dt}=\frac{2(t^2-1)}{(1+t^2)^2}$$

We want $$\int_{u=-1}^{u=1} \frac{2}{(u+2)\sqrt{1-u^2}}du = \int_{t=1}^{t=-1} \frac{2}{(u+2)(ut+1)} \cdot \frac{2(t^2-1)}{(1+t^2)^2} dt $$

and since $$(u+2)(ut+1) = \frac{-2t + 2(1 + t^2)}{1 + t^2} \cdot \frac{-2t^2 + (1 + t^2)}{1 + t^2} = \frac{2(1 - t + t^2)(1 - t^2)}{(1 + t^2)^2}$$

our integral simplifies to

$$\int_{-1}^{1} \frac{2}{1 - t + t^2} dt = \int_{-1}^{1} \frac{2}{(t-1/2)^2 + 3/4} dt $$

which can be written easily in terms of $\arctan$.

Alternatively, Euler's third substitution: $1-u^2$ has roots $\alpha = 1$ and $\beta = -1$. Putting $\sqrt{a(u - \alpha)(u - \beta)} = (u - \alpha)t$ gives, in our case, $\sqrt{1-u^2} = (u-1)t$, so $t = \frac{\sqrt{1-u^2}}{u-1}$. This rearranges to $u = \frac{t^2 - 1}{t^2 + 1}$ and so the derivative is $\frac{du}{dt} = \frac{4t}{(t^2 + 1)^2}$. Our integral becomes

\begin{align} \int_{-1}^1 \frac{2}{(u+2)\sqrt{1-u^2}}du &= \int_{t=0}^{t=-\infty} \frac{2}{(u+2)(u-1)t} \cdot \frac{4t}{(t^2 + 1)^2} dt \\ &= \int_{0}^{-\infty} \frac{2(t^2+1)^2}{(t^2 - 1 +2(t^2 + 1))(t^2 - 1 - 1(t^2 + 1))} \cdot \frac{4}{(t^2 + 1)^2} dt \\ &= \int_{-\infty}^{0} \frac{4}{3t^2 + 1} dt \\ &= \bigg[\frac{4 \arctan (\sqrt{3} t)}{\sqrt{3}} \bigg]_{-\infty}^0 \\ &= \frac{2\pi}{\sqrt{3}} \end{align}

(If you're wondering "which of these substitutions would Euler himself have chosen?" Well when Euler integrated $\int dx/(a + b \cos x)$ in 1768 he instead used the tangent half-angle substitution — long before Weierstrass, to whom that substitution is often misattributed, was born!)