I was given the following integral:
$$ \int \frac {x^2} {(3+4x-4x^2)} $$
By completing the square and using trigonometric substitution we get this:
$ \sqrt{-(4x^2-4x+1)+4 }=\sqrt{ -(2x-1)^2+4}$
So,
$2x-1=2sin\theta$
$x=sin\theta+\frac12$
$dx=\frac 12(2sin\theta+1)$
and
$\sqrt{ -(2x-1)^2+4}=\sqrt{(2sin\theta)^2+4}=2cos\theta$
$$ \int \frac {(sin\theta + \frac 12)^2cos\theta} {8cos^3\theta}d\theta $$
$$ \int \frac {sin^2\theta + sin\theta + \frac 14} {8cos\theta}d\theta $$
$$ \int \frac {sin^2\theta + sin\theta + \frac 14} {8cos^2\theta}d\theta $$
$$ \frac 18 \int \frac {sin^2\theta} {cos^2\theta}d\theta + \frac 18 \int \frac{sin\theta} {cos^2\theta}d\theta + \frac 1{32} \int \frac {1} {cos^2\theta}d\theta $$
From here there are two options involving the $ \frac 18 \int \frac {sin^2\theta} {cos^2\theta} $
The first, probably the more simple method, is substituting for $ tan^2\theta $ and subsequently $ sec^2\theta -1 $. Which looks like so:
$$ \frac 18 \int sec^2\theta d\theta - \frac 18 \int d\theta + \frac 18 \int tan\theta sec\theta d\theta + \frac 1{32} \int sec^2\theta $$
$$ \frac 5{32} \int sec^2\theta - \frac 18 \int d\theta + \frac 18 \int tan\theta sec\theta d\theta $$
$$ \frac 5{32} tan\theta - \frac 18 \theta + \frac 18 sec\theta + C $$
Substituting back in to get our original x values:
$$ \frac {5(2x-1)} {32\sqrt{-(2x-1)^2+4}} - \frac 18 sin^{-1}(\frac12(2x-1))+ \frac {2}{8\sqrt{-(2x-1)^2+4}} + C$$
$$ \frac {5(2x-1)} {32\sqrt{-(2x-1)^2+4}} + \frac {8}{32\sqrt{-(2x-1)^2+4}} - \frac 18 sin^{-1}(\frac12(2x-1)) + C$$
$$ \frac {10x-5} {32\sqrt{-(2x-1)^2+4}} + \frac {8}{32\sqrt{-(2x-1)^2+4}} - \frac 18 sin^{-1}(\frac12(2x-1)) + C$$
$$ \frac {10x+3} {32\sqrt{-(2x-1)^2+4}} - \frac 18 sin^{-1}(\frac12(2x-1)) + C$$
The other method I found was substituting $sin^2\theta$ for $1-cos^2\theta$ but I get a different solution.
$$ \frac 18 \int \frac {1-cos^2\theta} {cos^2\theta} + \frac 18 \int tan\theta sec\theta d\theta + \frac 1{32} \int sec^2\theta $$
$$ \frac 18 \int \frac {1} {cos^2\theta} d\theta - \frac 18 \int d\theta + \frac 18 \int tan\theta sec\theta d\theta \frac 1{32} \int sec^2\theta $$
$$ \frac 18 tan\theta - \frac 18 \theta + \frac 18 sec\theta + \frac 1{32}tan\theta + C $$
$$ \frac {2x-1} {8\sqrt{-(2x-1)^2+4}} + \frac {2}{8\sqrt{-(2x-1)^2+4}} + \frac {2x-1} {32\sqrt{-(2x-1)^2+4}} - \frac 18 sin^{-1}(\frac12(2x-1)) + C$$
$$ \frac {4(2x-1)} {32\sqrt{-(2x-1)^2+4}} + \frac {2}{8\sqrt{-(2x-1)^2+4}} + \frac {2x-1} {32\sqrt{-(2x-1)^2+4}} - \frac 18 sin^{-1}(\frac12(2x-1)) + C$$
$$ \frac {8x-4} {32\sqrt{-(2x-1)^2+4}} + \frac {2}{8\sqrt{-(2x-1)^2+4}} + \frac {2x-1} {32\sqrt{-(2x-1)^2+4}} - \frac 18 sin^{-1}(\frac12(2x-1)) + C$$
$$ \frac {10x-3} {32\sqrt{-(2x-1)^2+4}} - \frac 18 sin^{-1}(\frac12(2x-1)) + C$$
The solutions are nearly identical but the numerator in the first solution contains a +3 while the second contains a -3.
Note: I was given that the solution is:
$$ \frac {10x+3} {32\sqrt{-(2x-1)^2+4}} - \frac 18 sin^{-1}(\frac12(2x-1)) + C$$
In your second solution, you didn't convert the middle term to have a $32$ in the denominator (you left $\frac{2}{8\sqrt{\cdots}}$ instead of converting to $\frac{8}{32\sqrt{\cdots}}$).