I am asked to evaluate the following integral:

$\int_0^{2\pi} \cos^{10}\theta \mathrm{d}\theta$.

I am using complex analysis. Setting $z = e^{i\theta}$, I get from Eulers formula:

$\cos \theta = \frac{1}{2}\left(e^{i\theta} + e^{-i\theta}\right) = \frac{1}{2}\left(z + z^{-1}\right)$.

Now as $\theta$ goes from $0$ to $2\pi$, $z = e^{i\theta}$ goes one time around the unit circle. Therefore the problem is reduced to the following contour integral:

$\oint_{C} \left(\frac{1}{2}(z + z^{-1})\right)^{10} \frac{dz}{iz}$, where C is the unit circle.

At this point, I don't know how to move forward. I am pretty sure I am to apply the residue theroem, and the function I am integrating clearly has a pole at $z = 0$. But i don't know how to calculate the residue of that, since the pole is of the 10th order. Is there another approach i should take, maybe fine the laurent series of the function?

Any help is greatly appreciated!

  • $\begingroup$ Why the downvote ? $\endgroup$ – Surb Jul 27 at 14:47
  • $\begingroup$ Hint: $\frac{1}{2\pi i}\oint_{|z| = 1} \frac{dz}{z^k} = \begin{cases}1, & k = 1\\0, &k \in \color{red}{\mathbb{Z} \setminus \{ 1 \}}\end{cases}$ $\endgroup$ – achille hui Jul 27 at 15:52

Finding the residue of the meromorphic function $$ f(z):=\frac{(z+z^{-1})^{10}}{2^{10}iz} =\frac{1}{2^{10}i}\frac{(z^2+1)^{10}}{z^{11}}.\tag{1} $$ is not difficult.

But i don't know how to calculate the residue of that, since the pole is of the 10th order.

You are probably thinking about this formula for calculating residue at poles. But it is unnecessary here.

All you need is to find out the coeeficient of $z^{-1}$ in (1), which means you want the coefficient of $z^{10}$ in $(z^2+1)^{10}$. By the binomial theorem, one has $$ \frac{10!}{5!5!}=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6}{5\cdot 4\cdot 3\cdot 2}=9\cdot 4\cdot 7. $$ Hence the residue at $0$ is $$ \frac{63}{2^{8}i} $$ and by the residue theorem, the value of the integral is thus $$ 2\pi i\cdot \frac{63}{2^{8}i}=\frac{63\pi}{128}. $$

[Added:] Without complex analysis, one can still calculate the integral in just a few steps using the recursive formula of calculating $\int \cos^nx\,dx$ ($n>0$) and taking the advantage that we are integrating over the interval $[0,2\pi]$: $$ \begin{align} \int_{0}^{2\pi}\cos^{10}x\,dx &= \frac{9}{10}\int_{0}^{2\pi}\cos^{8}x\,dx \\ &= \frac{9}{10}\frac{7}{8}\int_{0}^{2\pi}\cos^{6}x\,dx\\ &= \frac{9}{10}\frac{7}{8}\frac{5}{6}\int_{0}^{2\pi}\cos^{4}x\,dx\\ &= \frac{9}{10}\frac{7}{8}\frac{5}{6} \frac{3}{4}\int_{0}^{2\pi}\cos^{2}x\,dx\\ &= \frac{9}{10}\frac{7}{8}\frac{5}{6} \frac{3}{4} \pi=\frac{63\pi}{128}. \end{align} $$

  • $\begingroup$ Great explanation on the complex analysis part. Thanks! $\endgroup$ – jakvah Jul 27 at 22:44


Set $$f(z)=\frac{(z+z^{-1})^{10}}{2^{10}iz}=\frac{(z^2+1)^{10}}{2^{10}iz^{11}}.$$

Then, $$f(z)=\frac{1}{iz^{11}}\sum_{k=0}^{10}\binom{10}{k}z^{2k}=...+\frac{1}{i2^{10}z}\binom{10}{5}+...$$

  • $\begingroup$ @jakvah: I forgot the 10 at the exponent. It should work now. $\endgroup$ – Surb Jul 27 at 15:02

Another method other than the residue theorem As $\cos^{10}{\theta}$ is an even function so $$\int_0^{2\pi} \cos^{10}{\theta} d\theta =2\int_0^{\pi} \cos^{10} \theta d\theta$$

Now let $\cos{\theta} =t$ $-\sin{\theta} d\theta = dt$

$\sin{\theta} = \sqrt{1-t^2}$

So integral become $$-2\int_1^{-1} \frac{t^{10}}{\sqrt{1-t^2}}dt$$ As the function formed in t is even, so by using the property

$\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx$

$$4\int_0^1 (t^{2})^5 (1-t^2) dt$$

Let $t^2=y$



So the integral become

$$2\int_0^1 y^\frac{9}{2} (1-y) dy$$

The beta function is given by $$\int_0^1 x^{m-1} (1-x)^{n-1} = \beta(m , n)$$

$m>0, n>0$

$$2\int y^\frac{9}{2} (1-y) dy = \beta(\frac{11}{2}, 2)$$

$m=\frac{11}{2}, n=2$

And, we know that $$\beta(m, n) = \frac{\Gamma{m}\Gamma{n}}{\Gamma{m+n}}$$


The value of integral is $$\int_0^{2\pi} \cos{\theta} d\theta = \frac{8}{2145}$$


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