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!

The problem asks to solve:

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

The answer is given as:

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

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?

  • $\begingroup$ Both answers are correct. $\endgroup$
    – Git Gud
    Feb 6, 2015 at 21:58
  • 1
    $\begingroup$ Thanks to everyone who answered! I greatly appreciate the help :) $\endgroup$
    – nomhtom
    Feb 6, 2015 at 22:14

2 Answers 2


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.


Your answer is also correct. Remember that antiderivates are only determined up to an additive constant.

$$\ln\left|{\sqrt{x^2+4}\over 2}+{x\over 2}\right|+C=\ln|\sqrt{x^2+4}+x|-\ln2+C.$$


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