Trig substitution problem. Stuck but close. I have this problem:

$$\int_0^2 \sqrt{1 + 4x^2}$$

Here's how I start. I focus on finding the antiderivative:
$$ \int \sqrt{4 * (\frac{1}{4} + x^2)}$$

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

so let $x = \frac{1}{2} * tan{\theta}$ so $dx = \frac{1}{2} * sec^2\theta d\theta$

so...

$$ \int 2 \sqrt{\frac{1}{4} + \frac{1}{4}tan^2{\theta}} * \frac{1}{2}sec^2{\theta} \cdot d\theta$$

and by $sec^2{\theta} = 1 + tan^2{\theta}$:

$$ \int 2 \cdot \frac{1}{2} sec{\theta} \cdot \frac{1}{2} sec^2{\theta} \cdot d\theta$$

$$ \int \frac{1}{2} sec{\theta} \cdot sec^2{\theta} \cdot d\theta$$ Let's call the above line : (1)

and now integration by parts:

$u = sec{\theta}$ and so $du = sec{\theta}tan{\theta}$

$dv = sec^2{\theta}$ and so $v = tan{\theta}$

so = $$\frac{1}{2} (\cdot sec{\theta}tan{\theta} - \int tan^2{\theta} sec{\theta} \cdot d\theta)$$

and so since $tan^2{\theta} = sec^2{\theta} - 1$

$$\frac{1}{2} (\cdot sec{\theta}tan{\theta} - \int sec^3{\theta} - \int sec{\theta} \cdot d\theta)$$

So using (1) we have:

$$(\frac{1}{2}  \cdot sec{\theta} \cdot tan{\theta} - \frac{1}{2} \int sec^3{\theta} - \frac{1}{2}  \int sec{\theta} \cdot d\theta) = \int \frac{1}{2} sec{\theta} \cdot sec^2{\theta} \cdot d\theta = \int \frac{1}{2} sec^3{\theta} d\theta$$

combining like sec^3{\theta} terms:

$$ (\frac{1}{2}  \cdot sec{\theta} \cdot tan{\theta} - \frac{1}{2}  \int sec{\theta} \cdot d\theta) = sec^3{\theta} $$

and now multiplying by $\frac{1}{2}$ again:

$$ (\frac{1}{4} \cdot sec{\theta} \cdot tan{\theta} - \frac{1}{4}  \int sec{\theta} \cdot d\theta) = \frac{1}{2} sec^3{\theta} $$

But what's the antiderivative of $\int sec{\theta} d\theta$ ?

I'm told to do this but I don't really get it:
\begin{align}
\int \sec \theta \, d\theta = \int \frac{\sec^2 \theta + \sec \theta \tan \theta}{\sec \theta + \tan \theta} \, d\theta=\ln (\sec \theta + \tan \theta) + c
\end{align}

Note that this is due to the numerator is the derivative of the denominator.

$$\int \frac{f'(t)}{f(t)} \, dt=\ln f(t) + c$$
Using a double-angle formula:

$$\sqrt{17}+\frac{1}{4} \sinh ^{-1}(4)$$