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?
