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.
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Calculus in IMO training courses? I thought IMO didn't go up to calculus – MT_ Apr 06 '14 at 02:28
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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 06 '14 at 08:17
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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. – MT_ Apr 06 '14 at 12:15
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@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 06 '14 at 16:27
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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.
André Nicolas
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1Did the $2007$ change to $2$? Sorry, I'm confused about the first step. Very nice solution though! – user139388 Apr 06 '14 at 02:18
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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 06 '14 at 02:19
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@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 06 '14 at 02:26
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@Jeff: Filling in should not be difficult. All that is being used is substitution. For the second substitution, we will need $\sin(\pi/2-u)=\cos u$ and $\cos(\pi/2-u)=\sin u$. – André Nicolas Apr 06 '14 at 02:29
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Can you add these details to your proof then. I will up vote & check answer if you can provide a full mathematical proof for the things you say in words. This is a math problem for IMO, and the solutions need to be detailed otherwise this won't get much credit at all... – Jeff Faraci Apr 06 '14 at 02:32
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I am confident there are only minor details to fill in. For example, when in $\frac{1}{2007^x+1}$ we let $x=-t$, then we get after a little algebra $\frac{2007^t}{1+2007^t}$. – André Nicolas Apr 06 '14 at 02:44
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1Naturally, if you point to an explicit place where you have had trouble, I will fill in any gap. – André Nicolas Apr 06 '14 at 03:13
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After getting integral $(1)$ it would have been best to direct put $x = \pi/2 - t$ and then by adding we would get $2I = \pi/2$. Splitting into $[0, \pi/4]$ and $[\pi/4, \pi/2]$ looks like a more lengthy route. – Paramanand Singh Apr 06 '14 at 04:22
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@ParamanandSingh: Yes, yours is a neater way of describing the symmetry. – André Nicolas Apr 06 '14 at 04:28
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@AndréNicolas See solution above to see what I was looking for. Thanks again for your solution to one of my problems however. – Jeff Faraci Apr 06 '14 at 07:58
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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 $$I_2=\int_0^{\dfrac\pi2}\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$
lab bhattacharjee
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