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 $\sin\theta$?

It doesn't need  any trigonometric substitution $$\int  \frac { x }{ \left( 1+x^{ 2 } \right) ^{ \frac { 3 }{ 2 }  } } dx=\frac { 1 }{ 2 } \int  \frac { d\left( 1+{ x }^{ 2 } \right)  }{ \left( 1+x^{ 2 } \right) ^{ \frac { 3 }{ 2 }  } } =\frac { 1 }{ 2 } \int { { \left( 1+{ x }^{ 2 } \right)  }^{ -\frac { 3 }{ 2 }  } } d\left( 1+{ x }^{ 2 } \right) =\\ =\frac { 1 }{ 2 } \frac { { \left( 1+{ x }^{ 2 } \right)  }^{ -\frac { 3 }{ 2 } +1 } }{ -\frac { 3 }{ 2 } +1 } =-\frac { 1 }{ \sqrt { 1+{ x }^{ 2 } }  } +C$$
