How to caluclate the integral of $\int \frac{1}{\sqrt{4x^{2}+1}}dx$ using a trig substitution? I am trying to determine the following integral: $\int \frac{1}{\sqrt{4x^{2}+1}} dx$ using a suitable substitution.
My progress:
let $x = \frac{1}{2} \tan \theta$
$dx = \frac{1}{2}\sec^{2} \theta d\theta$
$$\int \frac{1}{1+\tan^{2}\theta}\times \frac{1}{2}\times \sec^{2} \theta d\theta$$
$$= \frac{1}{2} \times \int 1 d\theta$$
$$ = \frac{1}{2} \theta + c$$
I know I have to get my answer back in terms of $x$, but I am a little stuck. Can someone please help?

You forgot the square root 
$$x=\frac{1}{2} \tan \theta \implies dx= \frac{1}{2} \sec^2 \theta d\theta $$
The integral becomes: 
$$\int \frac{1}{\sqrt{1+tan^2 \theta}} \cdot \frac{1}{2} \sec^2 \theta d\theta$$

$$\Large{\text{Not:}}$$
$$\bbox[8pt, border: crimson 4pt solid]{\int \frac{1}{1+tan^2 \theta} \cdot \frac{1}{2} \sec^2 \theta d\theta}$$
