Standard Integral Forms with Square Roots

Paralleling Section 5, the square-root forms $\int dx/\sqrt{a^2-x^2}$ and

$\int dx/\sqrt{x^2\pm a^2}$ are each derived once by a trigonometric substitution and then reused

for every quadratic radicand via completing the square.

Derivation of $\int dx/\sqrt{a^2-x^2}$ by trigonometric substitution. Let $x=a\sin\theta$,

$\theta\in(-\pi/2,\pi/2)$, so $dx=a\cos\theta\,d\theta$ and

$\sqrt{a^2-x^2}=\sqrt{a^2-a^2\sin^2\theta}=a\cos\theta$ (positive, since $\cos\theta\ge0$ on this

interval). Then

verified by $\dfrac{d}{dx}\sin^{-1}(x/a)=\dfrac{1/a}{\sqrt{1-x^2/a^2}}=\dfrac{1}{\sqrt{a^2-x^2}}$.

The other two square-root log forms (by the substitutions $x=a\tan\theta$ and

$x=a\sec\theta$ respectively, omitted here but directly checkable by differentiation):

A quadratic radicand $ax^2+bx+c$ is handled exactly as in Section 5 -- factor out $a$,

complete the square to $a[u^2\pm k^2]$ or $a[k^2-u^2]$ with $u=x+b/(2a)$ -- reducing to one of

the three forms above. Exercise: Standard Forms -- Square Roots applies this to

$\int dx/\sqrt{x^2+6x+13}$ and $\int(x+3)\,dx/\sqrt{x^2+6x+13}$ (numerator split exactly as in

Section 5, since the derivative of the radicand again equals a multiple of the numerator).

Integrating the square root itself: $\int\sqrt{a^2-x^2}\,dx$ and $\int\sqrt{x^2-a^2}\,dx$.

The same substitutions, now applied to the radical rather than its reciprocal, give (after using

the double-angle identity $\cos^2\theta=(1+\cos2\theta)/2$ -- carried out in full in Example 11):

$\int\sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\!\left(\frac{x}{a}\right)+C,$ …