How did people get the integral of 1 to be 3/2(x)?
I'm confused.
what is the integral of sin^2(x)?
Do you mean: $\displaystyle \int \left(\sin^{2} x + 2\sin x + 1 \right) dx$
First off, $\displaystyle \int 1 \: dx = \int 1x^{0} dx = \frac{x^{1}}{1} = x + c$
As for integrating $\displaystyle \sin^{2} x$, use the double angle identity: $\displaystyle \sin^{2} x = \frac{1 - \cos (2x)}{2}$
You made the substitution $\displaystyle x = \sin\theta + 1,\text{ where }-\frac\pi2\leq\theta\leq\frac\pi2$, yes?
So, we have:
$\displaystyle \theta = \arcsin(x - 1)$,
$\displaystyle \sin2\theta = 2\sin\theta\cos\theta\text{\quad\color{red}(trig identity)}$,
and
$\displaystyle \cos\theta = \sqrt{\cos^2\theta} = \sqrt{1 - \sin^2\theta}\text{\color{red}\quad(since the cosine is positive for our values of theta)}$
$\displaystyle \Rightarrow\cos\theta = \sqrt{1 - (x - 1)^2}$
Now just make the appropriate substitutions.