<s>[INST] 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?
 [/INST] Ultimately you want to re-express your answer $\frac{\tan\theta}{a^2}$ in terms of $\sin\theta$ and then convert back using the substitution you already did.  Observe that $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and that $\cos\theta = \sqrt{1 - \sin^2\theta}$.  Then
$$
\tan\theta \;\; =\;\; \frac{\sin\theta}{\cos\theta} \;\; =\;\; \frac{\sin\theta}{\sqrt{1 - \sin^2\theta}} \;\; =\;\; \frac{1}{a} \frac{x}{\sqrt{1 - \frac{x^2}{a^2}}} \;\; =\;\; \frac{x}{\sqrt{a^2 - x^2}}.
$$
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