Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
3  
Try comparing it to $\int(1/x^2)\,dx$. –  Gerry Myerson May 1 '12 at 12:11
2  
You have $1+\cot^2 u=\csc^2 u$ and $\cot^\prime u=-\csc^2 u$... –  J. M. 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

5 Answers 5

up vote 3 down vote accepted

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. . . .

share|improve this answer

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 $$\int_{0}^{\pi /2}\; f(x)\mathrm{d}x= \int_{0}^{\pi /2}\frac{1}{\tan ^{2}x}\mathrm{d}x.$$

--

$^{1}$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.

share|improve this answer

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.

Moreover, it it really not that hard to fill in the details in the above.

share|improve this answer

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 $$\int_0^{\pi/2} \frac{\sec^2(x)}{\tan^2(x)}dx$$ $$ $$

share|improve this answer

Wolfram|Alpha claims that no such integral exists:

http://tinyurl.com/7wt7p9p

Why do you ask?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.