# Integral $\int_{-\pi/2}^{\pi/2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x}dx$

I am trying to solve this integral $$\int_{-\pi/2}^{\pi/2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x}dx$$ A closed form does exist despite the looks of the integrand. This problem is from some old high school IMO training courses. I am not sure how to solve it. The only information that may be of help is $$\sin x=\sum_{n=1}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}, \quad \cos x=\sum_{n=1}^\infty \frac{ x^{2n}}{(2n)!} \quad \forall \ x$$ Thanks, I am looking for a complete solution, not a description as of what to do.

• Calculus in IMO training courses? I thought IMO didn't go up to calculus – MCT Apr 6 '14 at 2:28
• IMO is very advanced, just because they don't require calculus doesn't mean its not widely used in solutions!! @MichaelT most of the integrals I post are from mathematics high school (Plovdiv) and this is what we were doing preparing. You have to prepare for the most challenging problems in analysis, so if you cannot find the clever way to solve their problem, you can use the brute force methods. Have you taken the exam? If so, I know that you studied integration to the very day. – Jeff Faraci Apr 6 '14 at 8:17
• It is true that some IMO problems can be used solving calculus, but the trouble will be in knowing where to apply calculus, not how to solve a troublesome integral. – MCT Apr 6 '14 at 12:15
• @MichaelT This isn't that much of a troublesome integral at all relative to many. Many integrals are solved using methods such as pre-calculus or geometry. Often the calculus part can be small and in the background. – Jeff Faraci Apr 6 '14 at 16:27

Applying $\displaystyle\int_c^df(x)dx=\int_c^df(c+d-x)\ \ \ \ (1)dx$

$$I=\int_{-b}^b\frac1{a^x+1}\cdot\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$

$$=\int_{-b}^b\frac1{a^{-b+b-x}+1}\cdot\frac{\left(\sin(-b+b-x)\right)^{2n}}{\left(\cos(-b+b-x)\right)^{2n}+\left(\sin(-b+b-x)\right)^{2n}}dx$$

$$=\int_{-b}^b\frac{a^x}{a^x+1}\cdot\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$ as $\displaystyle\sin(-x)=-\sin x,\cos(-x)=\cos x$

$$\implies I+I=\int_{-b}^b\frac1{a^x+1}\cdot\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx+\int_{-b}^b\frac{a^x}{a^x+1}\cdot\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$

$$2I=\int_{-b}^b1\cdot\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$

Now setting $b=\dfrac\pi2,$

$$2I=\int_{-\dfrac\pi2}^{\dfrac\pi2}\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$

$$=\int_{-\dfrac\pi2}^0\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx +\int_0^{\dfrac\pi2}\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$

$$\text{For }I_1=\int_{-\dfrac\pi2}^0\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$

if $g(x)=\sin^{2n}x,g\left(-\dfrac\pi2+0-x\right)=\cdots=\cos^{2n}x$

using $(1),$ $$\text{For }I_1=\int_{-\dfrac\pi2}^0\frac{g(x)}{g\left(-\dfrac\pi2+0-x\right)+g(x)}dx=\int_{-\dfrac\pi2}^0\frac{g\left(-\dfrac\pi2+0-x\right)}{g(x)+g\left(-\dfrac\pi2+0-x\right)}dx$$

$$\implies I_1+I_1=\int_{-\dfrac\pi2}^0 dx=\cdots$$

Similarly, for $$I_2=\int_0^{\dfrac\pi2}\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$

• This is excellent and what I was looking for. Thanks a lot – Jeff Faraci Apr 6 '14 at 7:56

Separate into two parts, $-\pi/2$ to $0$ and $0$ to $\pi/2$.

For the integral from $-\pi/2$ to $0$, make the change of variable $t=-x$. We get $$\int_0^{\pi/2} \frac{2007^t}{2007^t+1} \frac{\sin^{2008}t}{\sin^{2008} t+\cos^{2008} t}\,dt.$$ Change the variable back to $x$, and add to the integral from $0$ to $\pi/2$.We get $$\int_0^{\pi/2}\frac{\sin^{2008}x}{\sin^{2008} x+\cos^{2008} x}\,dx.\tag{1}$$

Progress! Now split the integral (1) into the integral from $0$ to $\pi/4$, and the integral from $\pi/4$ to $\pi/2$.

For the integral from $\pi/4$ to $\pi/2$, let $u=\pi/2-x$, and use much the same trick we already used. The integral from $\pi/4$ to $\pi/2$ turns out to be $$\int_0^{\pi/4}\frac{\cos^{2008}u}{\cos^{2008} u+\sin^{2008} u}\,du.$$ Change the variable to $x$, and add to the integral from $0$ to $\pi/4$. We end up with $$\int_0^{\pi/4} 1\,dx,$$ which is not difficult.

• I am looking for a full complete solution. Thanks. – Jeff Faraci Apr 6 '14 at 2:12
• Did the $2007$ change to $2$? Sorry, I'm confused about the first step. Very nice solution though! – user139388 Apr 6 '14 at 2:18
• Yes, can you explain more of these steps? Many of the details are not here. I am looking for a full solution. It seems you are rushing to just post an answer that you know his correct, but are not explaining much. – Jeff Faraci Apr 6 '14 at 2:19
• @user139388: Thanks for pointing out the $2$ where $2007$ should be, I wasn't paying attention, since the actual number is irrelevant. (And I have a $10$ inch low res screen, so I tend to rely on memory instead of scrolling back.) – André Nicolas Apr 6 '14 at 2:26
• Naturally, if you point to an explicit place where you have had trouble, I will fill in any gap. – André Nicolas Apr 6 '14 at 3:13