# Integration of $\int_0^{\pi/2} \frac{1}{\tan^2(x)}dx$?

What is the value of the integration of $\int_0^{\pi/2} \frac{1}{\tan^2(x)}dx$?

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Try comparing it to $\int(1/x^2)\,dx$. –  Gerry Myerson May 1 '12 at 12:11
You have $1+\cot^2 u=\csc^2 u$ and $\cot^\prime u=-\csc^2 u$... –  Guess who it is. May 1 '12 at 12:13
The integral does not converge $-\cot x -x$ from $0$ to $\frac{\pi}{2}$ –  Kirthi Raman May 1 '12 at 12:18

Gerry Myerson hasn't made his into an answer and some time has passed, and no one else has posted such a simple answer. So here it is: Recall that $\tan x < 2x$ if $x$ is positive and close enough to $0$.

So \begin{align} \tan x & < 2x \\[12pt] \frac{1}{\tan^2 x} & > \frac 1 {4x^2} \\[12pt] \int_0^\bullet \frac{dx}{\tan^2 x} & \ge \int_0^\bullet \frac{dx}{4x^2} = \infty \end{align}

The number that appears where "$\bullet$" appears is not something we need to know precisely; just stare at the graph of the tangent function and you see that it works if "$\bullet$" is any sufficiently small positive number. If you want to be precise, notice that the derivative of the tangent function at $0$ is less than $2$ and stays less for a while as $x$ increases, etc. . . .

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This integral is an improper integral of the second kind (the singularity of the integrand is at $x=0$). The convergence or divergence of these integrals is normally proved by comparison with an integral whose convergence or divergence is known, either using the standard comparison test or the limit comparison test. The following is a proof that it does not converge.

Proof. Apply the limit comparison Test (LCT) for improper Integrals (Theorem 4.1). Let $f(x)=\dfrac{1}{\tan ^{2}x}$ and $g(x)=\dfrac{1}{x^{2}}$. Since

$$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0}\frac{x^{2}}{\tan ^{2}x}=1$$

and $$\int_{0}^{\pi /2}g(x)\; \mathrm{d}x=\int_{0}^{\pi /2}\frac{1}{x^{2}}\mathrm{d}x$$ is divergent$^{1}$, so is $$\int_{0}^{\pi /2}\; f(x)\mathrm{d}x= \int_{0}^{\pi /2}\frac{1}{\tan ^{2}x}\mathrm{d}x.$$

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$^{1}$We have $$\int_{a}^{b}\frac{1}{x^{2}}\mathrm{d}x=\left. -\frac{1}{x}\right\vert _{a}^{b}=\frac{b-a}{ab}.$$ Hence $$\int_{0}^{\pi /2}\frac{1}{x^{2}}\mathrm{d}x=\lim_{a\rightarrow 0}\int_{a}^{\pi /2} \frac{1}{x^{2}}\mathrm{d}x=\lim_{a\rightarrow 0}\frac{\pi /2-a}{a\pi /2}$$ is not finite.

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Here is an alternate solution

$$\int_0^{\pi/2} \frac{1}{\tan^2(x)}dx=\int_0^{\pi/2} \frac{\cos^2(x)}{\sin^2(x)}dx \geq \int_0^{\pi/2} \frac{\cos^3(x)}{\sin^2(x)}dx$$

Now, using $u= \sin(x)$

$$\int_0^{\pi/2} \frac{\cos^3(x)}{\sin^2(x)}dx= -\int_1^{0} \frac{1-u^2}{u^2}du=\int_0^{1} \frac{1}{u^2}du-1= \infty$$

P.S. Since $\lim_{x \to 0} sec(x) =1$, you can also compare your integral to $$\int_0^{\pi/2} \frac{\sec^2(x)}{\tan^2(x)}dx$$ 

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Wolfram|Alpha claims that no such integral exists:

http://tinyurl.com/7wt7p9p

Loosely speaking: At $x_0=0$ the function $x\mapsto\tan x$ behaves like $x\mapsto\sin x$ which behaves like $x\mapsto x$, hence the integral diverges.