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?
 A: 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

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

A: 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 $\sqrt{a^2-x^2}$. $$ \implies \tan \theta \ = \ \dfrac{\text{opposite}}{\text{adjacent}} \ = \ \dfrac{x}{\sqrt{a^2-x^2}}$$

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

A: 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
$$
\tan\theta \;\; =\;\; \frac{\sin\theta}{\cos\theta} \;\; =\;\; \frac{\sin\theta}{\sqrt{1 - \sin^2\theta}} \;\; =\;\; \frac{1}{a} \frac{x}{\sqrt{1 - \frac{x^2}{a^2}}} \;\; =\;\; \frac{x}{\sqrt{a^2 - x^2}}.
$$
A: We know that $\sin \theta=x/a$. From the identity $(\sin \theta)^2+(\cos\theta)^2=1$, we find $\cos\theta=\sqrt{1-(x/a)^2}$ (assuming $\theta$ is in the first quarter). From this you can find $\tan\theta=\sin \theta/\cos \theta$, which agrees with what the book says.
