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How can we evaluate $$\int_0^\pi\frac{1}{1+(\tan x)^\sqrt2}\ dx$$ Can you keep this at Calculus 1 level please? Please include a full solution if possible. I tried this every way I knew and I couldn't get it.

Sidharth Ghoshal
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Ovi
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    I like the title. Are you asking whether or not you need help? :-) – Asaf Karagila Apr 06 '13 at 23:01
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    Where did this integral come from? – Mhenni Benghorbal Apr 06 '13 at 23:24
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    That irrational power is a killer. Somehow I doubt a simple solution exists. – Mike Apr 07 '13 at 00:44
  • Maxima spits out some three screenfulls of a "simpler" integral including niceties like $e^{\cos x}$... – vonbrand Apr 07 '13 at 01:42
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    You say in a comment below that the bounds on the integral from $0$ to $\pi$. Omitting critically important information from your statement of a problem in your question is often yields unnecessary goose chases and wastes of other's efforts, so it is a courtesy to others to not hide things from them. – anon Apr 07 '13 at 02:47
  • @Ovi : I'll echo anon's point. If what you really want is a DEFINITE integral then you MUST include the limits of integration in the question. I'm surprised no one (including you) has edited the question. Sometimes a definite integral can be evaluated while the correpsonding indefinited integral can't be done. – Stefan Smith Apr 07 '13 at 15:28
  • I'm sorry, I didn't have the book with me when I posted this question so I didn't remember the limits of integration. I also didn't think it would make a difference. Until now I thought that the way you do all definite integrals is you find the indefinite integral and plug in the limits of integration, so once I had the indefinite I could plug in the numbers myself. Sorry again for the confusion this made. – Ovi Apr 07 '13 at 16:48
  • @Ovi : Quite often it is possible to evaluate a definite integral even when the antiderivative has no closed form. You could have mentioned that there were limits of integration that you had forgotten, and provided some probable values. – Stefan Smith Apr 07 '13 at 17:41

2 Answers2

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Indeed this was a Putnam question. The $\sqrt{2}$ is pretty much irrelevant. Notice

$$I=\int_0^\pi\frac{dx}{1+\tan(x)^{\sqrt2}}=\int_\pi^0\frac{d(\frac{\pi}{2}-u)}{1+(\tan(\frac{\pi}{2}-u))^{\sqrt2}}=\int_0^\pi\frac{du}{1+\cot(u)^{\sqrt2}}$$

and

$$\frac{1}{1+\tan(x)^{\sqrt2}}=\frac{\cos(x)^{\sqrt2}}{\cos(x)^{\sqrt2}+\sin(x)^{\sqrt2}},\qquad\frac{1}{1+\cot(u)^{\sqrt2}}=\frac{\sin(u)^{\sqrt2}}{\sin(u)^{\sqrt2}+\cos(u)^{\sqrt2}}.$$

so

$$2I=\int_0^\pi\frac{dv}{1+\tan(v)^{\sqrt2}}+\int_0^\pi\frac{dv}{1+\cot(u)^{\sqrt2}}$$

$$=\int_0^\pi\left[\frac{\cos(v)^{\sqrt2}}{\cos(v)^{\sqrt2}+\sin(v)^{\sqrt2}}+\frac{\sin(v)^{\sqrt2}}{\sin(v)^{\sqrt2}+\cos(v)^{\sqrt2}}\right]dv=\int_0^\pi\frac{\cos(v)^{\sqrt2}+\sin(v)^{\sqrt2}}{\cos(v)^{\sqrt2}+\sin(v)^{\sqrt2}}dv $$

which is $\int_0^\pi1dv=\pi$. Ultimately, this is a symmetry argument.

anon
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This is a Putnam problem from years ago. There is no Calc I solution of which I'm aware. You need to put a parameter (new variable) in place of $\sqrt 2$ and then differentiate the resulting function of the parameter (this is usually called "differentiating under the integral sign"). Most students don't even learn this in Calc III!

Ted Shifrin
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    Differentiation under the integral sign for indefinite integrals? O_o – Foahen Apr 07 '13 at 01:51
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    Oops. My fault. The original Putnam problem was the integral from $0$ to $\pi/2$. The problem as stated here must be undoable. – Ted Shifrin Apr 07 '13 at 02:05
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    I'm confused. if it is a definite integral from 0 to $\pi/2$, simple change of variable $x \to \frac{\pi}{2} - x$ is enough to get the answer $\pi/4$. Why do we need to use "differentiate under the integral sign" trick? – achille hui Apr 07 '13 at 02:24
  • Yes, I didn't think it was important to include the boundaries but in my book it is from 0 to pi/2. However, I am only in AP calculus AB (which is equivalent to calculus 1) and this is in the chapter about the integrals and derivatives of logarithmic and inverse trig functions – Ovi Apr 07 '13 at 02:33
  • So can you show the steps to using this x->pi/2-x trick? I've never seen it before. – Ovi Apr 07 '13 at 02:35
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    @Ovi, under $x \to \frac{\pi}{2} - x$, $\tan x \to \frac{1}{\tan x}$ and $\frac{1}{1 + (\tan x)^{\alpha}} \to \frac{(\tan x)^{\alpha}}{1 + (\tan x)^{\alpha}}$. So $$\int_{0}^{\pi/2} \frac{1}{1 + (\tan x)^{\alpha}} dx = \int_{0}^{\pi/2} \frac{(\tan x)^{\alpha}}{1 + (\tan x)^{\alpha}} dx$$ Adding both sides and divide by two, you get $\int_{0}^{\pi/2} \frac12 dx = \pi/4$. – achille hui Apr 07 '13 at 02:49