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.

Help me please to compute: $\int_{2}^{3}\sqrt{1+\frac{1}{x^{2}}} \: \: dx$

Thanks a lot!

share|improve this question
add comment

3 Answers 3

up vote 3 down vote accepted

Hint: let $x=\tan{\theta}$, $dx=\sec^2{\theta} d \theta$:

$$\begin{align} \int_{2}^{3} dx \: \sqrt{1+\frac{1}{x^{2}}} &= \int_{\arctan{2}}^{\arctan{3}} \frac{d \theta}{\sin{\theta} \cos^2{\theta}} \\ &= \int_{\arctan{2}}^{\arctan{3}} \frac{d \theta \, \sin{\theta}}{\sin^2{\theta} \cos^2{\theta}} \end{align} $$

Now let $y=\cos{\theta}$, $dy=-\sin{\theta} d \theta$:

$$\begin{align} \int_{2}^{3} dx \: \sqrt{1+\frac{1}{x^{2}}} &= \int_{\frac{1}{\sqrt{10}}}^{\frac{1}{\sqrt{5}}} \frac{dy}{y^2 (1-y^2)} \\ &= \int_{\frac{1}{\sqrt{10}}}^{\frac{1}{\sqrt{5}}} dy \: \left ( \frac{1}{y^2} + \frac{1}{1-y^2} \right )\\ &= \int_{\frac{1}{\sqrt{10}}}^{\frac{1}{\sqrt{5}}} dy \: \left [ \frac{1}{y^2} + \frac{1}{2} \left ( \frac{1}{1-y} + \frac{1}{1+y} \right ) \right ] \\ \end{align} $$

I think you can take it from here.

share|improve this answer
add comment

We let $x=1/y$ that yields $$-\int_{1/2}^{1/3}\frac{\sqrt{1+y^2}}{y^2} dy=$$ $$\int_{1/2}^{1/3}\left(\frac{1}{y}\right)'\sqrt{1+y^2} dy=$$ $$\left[\frac{\sqrt{1+y^2}}{y}\right]_{1/2}^{1/3}-\int_{1/2}^{1/3} \frac{1}{\sqrt{1+y^2}} dy=\sqrt{10}-\sqrt{5} -\sinh^{-1}(1/3)+\sinh^{-1}(1/2)$$

share|improve this answer
add comment

Hint : set $x = f(y)$ where is $f$ is a very well-known function.

share|improve this answer
    
Not informative and questionable hint. For example $x=sin(y)$ and it gives no way out in finding the solution. Could you please elaborate more on how your hint could/should help? –  Tomas Jan 28 '13 at 12:50
add comment

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.