Need Clarification on Trig Substitution $$\int\frac{x}{(1+x^2)^\frac{3}{2}}\ \mathrm dx$$
becomes 
$$\sin\theta = \frac{x}{\sqrt{x^2+1}}$$
I am aware that we would use $x = a^2\tan^2\theta$. and that our answer before we plug in from the triangle becomes $\sin\theta$.
I just took a test and got the answer of 
$$\sin\theta = \frac{x}{(x^2+1)^\frac{3}{2}}$$
Why does the $\dfrac{3}{2}$ become  $\dfrac{1}{2}$ when we are plugging in for $\sin\theta$?

I'd write $\displaystyle \int \frac 1 {(x^2+1)^{3/2}} \Big( x\, dx\Big) = \int \frac 1 {u^{3/2}} \left( \frac 1 2\,du\right) = \cdots,$ etc.
But if you must use a trigonometric substitution you can do this:
\begin{align}
x & = \tan\theta \\
dx & = \sec^2\theta\,d\theta \\[10pt]
\int \frac x {(1+x^2)^{3/2}} \,dx & = \int \frac{\tan\theta}{\sec^3\theta} \, \sec^2\theta\,d\theta = \int \sin\theta\,d\theta = -\cos\theta + C \\[10pt]
& = \frac{-1}{\sqrt{1+x^2}} + C
\end{align}
