# Integrating $\int \frac{1}{\sqrt{x^2+4}}\,dx$ using Trigonometric Substitution

I'm reviewing integration by trigonometric substitution in anticipation of covering it in class next week. I seem to be a bit rusty and keep catching myself making various mistakes. On this particular problem I keep getting the same answer which is very close to being correct. However, I somehow end up dividing by two where I should not. I'm hoping another set of eyes can quickly set me right so I can stop frustrating myself reworking the problem to the same apparently wrong answer repeatedly! Thanks in advance!

$$\int \frac{1}{\sqrt{x^2+4}}\,dx$$

$$ln\lvert x + \sqrt{x^2+4} \rvert + C$$

Somehow I keep getting:

$$ln\Bigg|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\Bigg| + C$$

Here's my work:

$$\int \frac{1}{\sqrt{x^2+4}}\,dx = \int \frac{1}{\sqrt{4(\frac{1}{4}x^2+1)}}\,dx = \frac{1}{2}\int\frac{1}{\sqrt{\frac{1}{4}x^2+1}}\,dx = \frac{1}{2}\int\frac{1}{\sqrt{(\frac{1}{2}x)^2+1}}\,dx$$

At this point I substitute as follows:

$$\frac{1}{2}x = \tan\theta$$ $$x = 2\tan\theta$$ $$dx = 2\sec^2\theta$$

So I continue on with:

$$\frac{1}{2}\int\frac{2\sec^2\theta}{\sqrt{\tan^2\theta+1}}\,d\theta = \int\frac{\sec^2\theta}{\sqrt{\sec^2\theta}}\,d\theta = \int\frac{\sec^2\theta}{\sec\theta}\,d\theta = \int\sec\theta\,d\theta = ln\lvert\sec\theta + \tan\theta\rvert + C$$

Finally, to get the answer in terms of x I essentially draw a right triangle and use the fact that $\tan\theta = \frac{x}{2}$. The side opposite $\theta$ I take to be x, the side adjacent $\theta$ is 2, and the hypotenuse is $\sqrt{x^2+4}$. So $\sec\theta = \frac{\sqrt{x^2+4}}{2}$ and $\tan\theta = \frac{x}{2}$.

So, substituting these values back in, as mentioned, I end up with:

$$ln\Bigg|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\Bigg| + C$$

Can anyone help me see where I'm going wrong or failing to understand something?

• Both answers are correct. Feb 6, 2015 at 21:58
• Thanks to everyone who answered! I greatly appreciate the help :) Feb 6, 2015 at 22:14

You are correct still. Notice that \begin{align} \ln\Bigg|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\Bigg| + C &= \ln\Bigg|\frac{\sqrt{x^2+4}+x}{2}\Bigg| + C\\ &= \ln\Bigg|\sqrt{x^2+4}+x\Bigg|-\ln(2) + C\\ &= \ln\Bigg|\sqrt{x^2+4}+x\Bigg|+C' \end{align} where $C'$ is still an arbitrary constant.
$$\ln\left|{\sqrt{x^2+4}\over 2}+{x\over 2}\right|+C=\ln|\sqrt{x^2+4}+x|-\ln2+C.$$