# Solve trigonometric integral $\int_{-\pi/2}^{\pi/2} \frac{\sin^{2014}x}{\sin^{2014}x+\cos^{2014}x} dx$

Please help me to solve the following integral: $$\int_{-\pi/2}^{\pi/2} \frac{\sin^{2014}x}{\sin^{2014}x+\cos^{2014}x} dx$$ I have tried a lot, but no results. I only transformed this integral to the following also not easy integral: $$\int \frac{1}{(1+t^2)(1+t^{2014})} dt.$$

• Use the fact that $\sin(\frac{\pi}2-x)=\cos x$. May 30 '15 at 16:59
• ... or split the last integral over $[0,1]$ and $[1,+\infty)$, then apply the change of variable $t\to\frac{1}{s}$ in the second integral. May 30 '15 at 17:04
• Jack D'Aurizio, I tried your change, it didn't help May 30 '15 at 17:11
• Just a remark: I have used Mathematica to show that the function $F(a) = \int^{\pi/2}_{-\pi/2} \frac{\sin^a{x}}{\sin^a{x} + \cos^a{x}} \, \mathrm{d}x$ has the following values: $F(0)=F(2)=F(4)=F(6)=F(8) = \pi/2$ and $F(1), \ F(3), \, F(5)$ and $F(7)$ are not convergent in $(-\pi/2,\pi/2)$. Maybe you can compute easily what $F(2k)$ is. May 30 '15 at 17:11
• This boils down to this general integral: math.stackexchange.com/questions/605673/… May 30 '15 at 19:34

Consider the integral \begin{align} I = \int_{-\pi/2}^{\pi/2} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx. \end{align} This may also be seen as \begin{align} I &= \int_{0}^{\pi/2} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx + \int_{-\pi/2}^{0} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx \\ &= \int_{0}^{\pi/2} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx - \int_{\pi/2}^{0} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx \\ &= 2 \, \int_{0}^{\pi/2} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx \end{align} where $x \to -x$ was made in the second integral. Now make the substitution $x = t-\pi/2$ to obtain \begin{align} I = 2 \, \int_{-\pi/2}^{0} \frac{\cos^{2a}t}{\sin^{2a}t + \cos^{2a}t} \, dt. \end{align} Now let $t \to -x$ to obtain \begin{align} I = 2 \, \int_{0}^{\pi/2} \frac{\cos^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx. \end{align} Adding the two integral expressions leads to \begin{align} 2 I &= 2 \, \int_{0}^{\pi/2} \frac{\sin^{2a}x + \cos^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx = 2 \, \int_{0}^{\pi/2} dx = \pi. \end{align} It can now be stated that \begin{align} \int_{-\pi/2}^{\pi/2} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx = \frac{\pi}{2}. \end{align}

• You were way faster than me! Very nice, by the way. May 30 '15 at 17:32
• why $sin^{2a}(-x)=sin^{2a}(x)$, why is even function? May 30 '15 at 17:52
• $\sin^{2a}(-x) = (-1)^{2a} \, \sin^{2a}(x) = \sin^{2a}(x)$ since $(-1)^{2m} = ((-1)^{2})^{m} = (1)^{m} = 1$ May 30 '15 at 17:58
• what did you led to use this method? is a general method or just an inspiration? May 30 '15 at 18:01
• @Lucas Problems of this type sometimes show up in the Putnam exam, IMO, USAMO, and other exam competitions. Calculus books from the late 1800's to 1930's sometimes use problems of this type as examples. Outside of that it is just experience that often finds nice solutions. May 30 '15 at 18:31

Notice the following properties of limits, $$\color{blue}{\int_{-a}^{a}f(x)dx =2\int_{0}^{a}f(x)dx\quad \text{if} \quad f(-x)=f(x)}$$ $$\color{blue}{\int_{0}^{a}f(x)dx =\int_{0}^{a}f(a-x)dx}$$

Now, let $$I=\int_{-\pi/2}^{\pi/2}\frac{\sin^{2014}x}{\sin^{2014}x+\cos^{2014}x}dx$$ $$\implies I=2\int_{0}^{\pi/2}\frac{\sin^{2014}x}{\sin^{2014}x+\cos^{2014}x}dx\tag 1$$ Using secong property, we get $$I=2\int_{0}^{\pi/2}\frac{\sin^{2014}\left(\frac{\pi}{2}-x\right)}{\sin^{2014}\left(\frac{\pi}{2}-x\right)+\cos^{2014}\left(\frac{\pi}{2}-x\right)}dx$$

$$\implies I=2\int_{0}^{\pi/2}\frac{\cos^{2014}x}{\sin^{2014}x+\cos^{2014}x}dx\tag 2$$ Now, adding eq(1) & (2), we get $$2I=2\int_{0}^{\pi/2}\left(\frac{\sin^{2014}x}{\sin^{2014}x+\cos^{2014}x}+\frac{\cos^{2014}x}{\sin^{2014}x+\cos^{2014}x}\right)dx$$ $$\implies I=\int_{0}^{\pi/2}\left(\frac{\sin^{2014}x+\cos^{2014}x}{\sin^{2014}x+\cos^{2014}x}\right)dx$$ $$=\int_{0}^{\pi/2}1. dx=[x]_{0}^{\pi/2}=\left[\frac{\pi}{2}-0\right]=\color{blue}{\frac{\pi}{2}}$$

Another point of view, just to point out what's going on with the integral:

Consider the integral written in the form:

$$I(2k) = \int^{\pi/2}_{-\pi/2} \frac{1}{1+\cot^{2k}{x}} \, \mathrm{d} x,$$ and look at what happens with the integrand as $k$ goes to $\infty$ (in your case, $k = 512$):

where I have ranged $k$ from $2$ to $20$. Can you infer an asymptotic behavior of $I(2k)$ as $k \to \infty$?

Cheers!

Edit: I was fascinated by the fact pointed out by @enzotib. In fact, this may suggest that $\partial I/\partial k = 0$. Let's compute the derivative of the integrand w.r.t. $k$:

$$\frac{\partial}{\partial k} \frac{1}{1+\cot^{2k}{x}} = -\frac{2 \cot ^{2 k}(x) \log (\cot (x))}{\left(\cot ^{2 k}(x)+1\right)^2} =:f.$$ We observe that $f$ is only defined for $x > 0$ and, as a matter of fact, it's an odd function with respect the line $x = \pi/4$ for all $k$. Therefore, $F= \int^{\pi/2}_{-\pi/2} f \, \mathrm{d} x= 0$ for every $k$, meaning that $I$ does not depend on $k$. Since $F = \partial I / \partial k$, on integrating $F$ with respect to $k$, we can see that $I = c$, for some constant $c$. To determine the value of $c$, we plug any value of $k$ in $I$ to have that $I = I(0) = \int^{\pi/2}_{-\pi/2} \mathrm{d}x /2 = \pi/2$.

• Very nice geometric way to see this. +1 May 30 '15 at 19:03
• Most important: the area under the curve does not change with $k$. May 30 '15 at 22:36
• I came around to this question by chance and edited my post to prove your claim, @enzotib; in an empirical way though. Jun 20 '15 at 3:39

For $$\int_a^{\pm(\pi/2-a)} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx$$

I will use $$I=\int_p^q f(x)\ dx=\int_p^q f(p+q-x)\ dx$$

$$\implies I+I=\int_p^q [f(x)+ f(p+q-x)]\ dx$$