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: $$\int \frac{1}{(1+t^2)(1+t^{2014})} dt.$$

Solution 1:

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}

Solution 2:

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

Solution 3:

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 $I = I(0) = \int^{\pi/2}_{-\pi/2} \mathrm{d}x /2 = \pi/2$.

Solution 4:

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

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

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