Expressing the integral in terms of the original variable In evaluating the integral:
$$ \int{dx\over(a^2-x^2)^{3/2}} $$ or $$ \int{dx\over(a^2-x^2)^{1/2}\ (a^2-x^2)}$$ 
Let $ x=a\sin\theta $ and $ dx=a\cos\theta\ d\theta $. Then
$$ \int{{a\cos\theta\ d\theta}\over{a\cos\theta\ (a^2-a^2\sin^2\theta)}} = {1\over a^2}\int {{d\theta}\over{1-\sin^2\theta}} = {1\over a^2}\int {{d\theta}\over{\cos^2\theta}} = {1\over a^2}\int \sec^2\theta\ d\theta $$
$$ ={{\tan \theta} \over {a^2}}$$
My question has to do with the method of expressing the transformed variable $\theta$ in terms of the original variable $x$. Taking the initial substitution $x=a\sin\theta$, I am able to derive the following:
$$ \theta=\arcsin{x \over a}$$
And substitute into the answer in the following manner:
$$ \tan(\arcsin {x\over a})\over a^2$$
However, the book expresses the integral as such:
$$ {1 \over a^2}{{x} \over {\sqrt {a^2-x^2}}} + C$$
Although the term $1 \over a^2$ remains the same, how do I express the trigonometric functions in this particular form?

Solution without trigonometric functions.
Given

$$
\int \frac{dx}{ \sqrt{ a^2 - x^2 }^3 }.
$$

Write this as

$$
\int \frac{dx}{ \sqrt{ a^2 - x^2 }^3 }
= \frac{1}{a^2} \int \frac{a^2 - x^2 + x^2}{ \sqrt{ a^2 - x^2 }^3 } dx
$$

or

$$
\int \frac{dx}{ \sqrt{ a^2 - x^2 }^3 }
= \frac{1}{a^2} \int \frac{dx}{ \sqrt{ a^2 - x^2 } }
  + \color{blue}{ \frac{1}{a^2} \int \frac{x^2}{ \sqrt{ a^2 - x^2 }^3 } dx }.
$$

By parts on the blue part gives

$$
\begin{eqnarray}
\int \frac{d x}{ \sqrt{ a^2 - x^2 }^3 }
&=& \frac{1}{a^2} \int \frac{d x}{ \sqrt{ a^2 - x^2 } }
     + \color{blue}{ \frac{1}{a^2} \int x \frac{x}{ \sqrt{ a^2 - x^2 }^3 } dx }\\
&=& \color{red} { \frac{1}{a^2} \int \frac{d x}{ \sqrt{ a^2 - x^2 } } }
     + \bbox[16px,border:2px solid #800000] { \frac{1}{a^2} \frac{x}{ \sqrt{ a^2 - x^2 } } }
     - \color{red} { \frac{1}{a^2} \int \frac{d x}{ \sqrt{ a^2 - x^2 } } }.
\end{eqnarray}
$$

So the final result is

$$
\int \frac{d x}{ \sqrt{ a^2 - x^2 }^3 } = \frac{1}{a^2} \frac{x}{ \sqrt{ a^2 - x^2 } }.
$$

