# Difficult integration

• May 15th 2011, 11:42 AM
jackie
Difficult integration
I was helping my friend with her integration problems, and both she and I got stuck on this problem. Anyone can gives me a hand. I'd really appreciate it. We were trying to integrate $\frac{\sqrt(x^2+1)}{x^2}$. We tried to do the following substitution. Let $x=tan\theta$ then, $dx=sec^2\theta$. So we have $\int \frac{(tan^2\theta +1)sec^2\theta}{tan^2\theta}=\int \frac{sec^3\theta}{tan^2\theta}$. Then since $tan^2\theta=sec^2\theta-1$, we substitute this into the expression, but got nowhere.
• May 15th 2011, 11:56 AM
HallsofIvy
For problems like these, I almost always throw my hands in the air and convert to sine and cosine! $sec(\theta)= \frac{1}{cos(\theta)}$ and $tan(\theta)= \frac{sin(\theta)}{cos(\theta)}$ so $\frac{sec^3(\theta)}{tan^2(\theta)}$ $= \frac{1}{cos^3(\theta)}\frac{cos^2(\theta)}{sin^2( \theta)}$ $= \frac{1}{cos(\theta)sin^2(\theta)}$.

Now, that has an odd power of cosine, even though it is in the denominator, so I would multiply both numerator and denominator by $cos(\theta)$ in order to get
$\frac{cos(\theta)}{cos^2(\theta)sin^2(\theta)}= \frac{cos(\theta)}{(1- sin^2(\theta))\sin(\theta)}$. Make the substitution $u= sin(\theta)$ so that $du= cos(\theta)d\theta$ and that integral becomes
$\frac{du}{u(1-u^2)}$
which can be integrated by partial fractions.
• May 15th 2011, 04:32 PM
Soroban
Hello, jackie!

Quote:

$\text{Integrate: }\:\int\frac{\sqrt{x^2+1}}{x^2}\,dx$

$\text{We tried this substitution: }\:x=\tan\theta \quad\Rightarrow\quad dx=\sec^2\!\theta\,d\theta$

$\text{So we have: }\:\int \frac{\sqrt{\tan^2\!\theta +1}\sec^2\!\theta}{\tan^2\!\theta}\;=\;\int \frac{\sec^3\!\theta}{\tan^2\!\theta}\,d\theta$

$\text{Then since }\tan^2\!\theta\,=\,\sec^2\!\theta-1,\,\text{ we substituted this, but got nowhere.}$

I did it like this . . .

. . $\frac{\sec^3\!\theta}{\tan^2\!\theta} \;=\;\frac{\sec\theta\cdot\sec^2\!\theta}{\tan^2\! \theta} \;=\;\frac{\sec\theta(\tan^2\!\theta + 1)}{\tan^2\!\theta} \;=\;\sec\theta + \frac{\sec\theta}{\tan^2\!\theta}$

. . . $=\;\sec\theta + \frac{1}{\cos\theta}\!\cdot\!\frac{\cos^2\!\theta} {\sin^2\!\theta} \;=\; \sec\theta + \frac{\cos\theta}{\sin^2\!\theta}$

. . . $=\;\sec\theta + \frac{1}{\sin\theta}\cdot\frac{\cos\theta}{\sin \theta} \;=\;\sec\theta + \csc\theta\cot\theta$

$\text{The integral becomes:}$

. . $\int (\sec\theta + \csc\theta\cot\theta)\,d\theta \;=\; \ln|\sec\theta + \tan\theta| - \csc\theta + C$

$\text{Back-substitute: }\:\tan\theta \,=\,x,\;\;\sec\theta \,=\,\sqrt{x^2+1},\;\;\csc\theta \,=\,\frac{\sqrt{x^2+1}}{x}$

$\text{Therefore: }\;\ln\left|\sqrt{x^2+1} + x\right| - \frac{\sqrt{x^2+1}}{x} + C$

• May 15th 2011, 08:10 PM
jackie
Thanks a lot for your help, HallsofIvy and Soroban. My friend and I followed HallsofIvy's hint and got the answer. Later today we see Soroban's answer, and my friend likes the way you did this problem. Anyway, we were glad to do this problem with both ways. Really appreciate both of your help.