what is integral of sqrt(1+4x^2)?
I tried looking at the trig rules but it was only if that part was in the denominator ... how would I find the integral of that?
we will use trigonometric substitution:
$\displaystyle \int \sqrt {1 + 4x^2}~dx$
We proceed by trig substitution:
Let $\displaystyle 2x = \tan \theta $
$\displaystyle \Rightarrow 2~dx = \sec^2 \theta ~d \theta$
$\displaystyle \Rightarrow dx = \frac {1}{2} \sec^2 \theta ~d \theta$
So our integral becomes
$\displaystyle \frac {1}{2} \int \sec^2 \theta \cdot \sec \theta~d \theta$
Now continue, remember to back substitute when finished to get the answer in terms of $\displaystyle x$
$\displaystyle \int\sqrt{1+4x^2}\,dx=\frac12\int\sqrt{1+4\left(\f rac{\tan^2\theta}4\right)}\sec^2\theta\,d\theta=\f rac12\int\sec\theta\sec^2\theta\,d\theta$
Where $\displaystyle 1+\tan^2\theta=\sec^2\theta$
Is it clearer now?
You can do the remaining integral using integration by parts.