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?

You can form a right angled triangle from your substitution $x = a\sin\theta$. This'll give the hypotenuse as $a$, opposite side as $x$ and by Pythagoras' theorem, your adjacent side will have length $\sqrt{a^2-x^2}$. $$ \implies \tan \theta \ = \ \dfrac{\text{opposite}}{\text{adjacent}} \ = \ \dfrac{x}{\sqrt{a^2-x^2}}$$

$$ \therefore \ \int \dfrac{1}{\left(a^2-x^2\right)^{3/2}} \text{ d}x \ = \ \dfrac{1}{a^2} \dfrac{x}{\sqrt{a^2-x^2}} + \mathcal{C} $$

