Some Special Integrals

The seven special integrals listed above are each proved by a trigonometric substitution, chosen according to which quadratic surd or quadratic denominator is present (see the substitution table).

Formula 1: $\int\dfrac{dx}{x^2+a^2}=\dfrac1a\tan^{-1}\dfrac xa+c$. Put $x=a\tan\theta$, so $\theta=\tan^{-1}\frac xa$ and $dx=a\sec^2\theta\,d\theta$. Then $x^2+a^2=a^2\sec^2\theta$, so the integral becomes $\int\dfrac{a\sec^2\theta}{a^2\sec^2\theta}\,d\theta=\dfrac1a\int d\theta=\dfrac1a\theta+c=\dfrac1a\tan^{-1}\dfrac xa+c$. (This can also be checked directly: differentiating $\frac1a\tan^{-1}\frac xa$ gives $\frac1a\cdot\frac{1}{1+x^2/a^2}\cdot\frac1a=\frac{1}{x^2+a^2}$, confirming the formula without any substitution at all.)

Formulas 2 & 3. $\int\dfrac{dx}{x^2-a^2}$: factor $x^2-a^2=(x-a)(x+a)$ and split into partial fractions, $\frac{1}{x^2-a^2}=\frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right)$; integrating each piece gives $\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+c$. The companion result $\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+c$ follows by the identical partial-fraction split with the sign of $x^2$ reversed.

Formula 4: $\int\dfrac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\dfrac xa+c$. Put $x=a\sin\theta$, so $\theta=\sin^{-1}\frac xa$, $dx=a\cos\theta\,d\theta$, and $\sqrt{a^2-x^2}=a\cos\theta$. The integral collapses to $\int d\theta=\theta+c=\sin^{-1}\frac xa+c$.

Formula 5: $\int\dfrac{dx}{\sqrt{x^2-a^2}}=\log\left|x+\sqrt{x^2-a^2}\right|+c$. Put $x=a\sec\theta$, $dx=a\sec\theta\tan\theta\,d\theta$, and $\sqrt{x^2-a^2}=a\tan\theta$; the integral reduces to $\int\sec\theta\,d\theta=\log|\sec\theta+\tan\theta|+c_1$. Converting back to $x$ (using $\sec\theta=x/a$, $\tan\theta=\sqrt{x^2-a^2}/a$) and absorbing $-\log a$ into the constant gives the stated formula. Formula 6 ($\int\frac{dx}{\sqrt{x^2+a^2}}=\log|x+\sqrt{x^2+a^2}|+c$) is obtained the same way using $x=a\tan\theta$.

Formula 7: $\int\dfrac{dx}{x\sqrt{x^2-a^2}}=\dfrac1a\sec^{-1}\dfrac xa+c$. Put $x=a\sec\theta$ as before; $x\sqrt{x^2-a^2}=a\sec\theta\cdot a\tan\theta$, and $dx=a\sec\theta\tan\theta\,d\theta$, so the $\sec\theta\tan\theta$ factors cancel, leaving $\frac1a\int d\theta=\frac1a\theta+c=\frac1a\sec^{-1}\frac xa+c$.

These formulae are most powerful once combined with a substitution that first turns a more complicated integrand into one of these seven shapes:

$\int\dfrac{\sin2x}{3\sin^4x-4\sin^2x+1}\,dx$. Put $\sin^2x=t$, so $2\sin x\cos x\,dx=\sin2x\,dx=dt$: integral becomes $\int\dfrac{dt}{3t^2-4t+1}$. Completing the square (see 3.2.4) gives $\frac13\int\dfrac{dt}{\left(t-\frac23\right)^2-\left(\frac13\right)^2}$, a Formula-2 shape, and the final answer works out to $\dfrac12\log\left|\dfrac{3\sin^2x-3}{3\sin^2x-1}\right|+c$.

$\int\dfrac{e^{x/2}}{\sqrt{e^{-x}-e^x}}\,dx$. Simplify the integrand to $\dfrac{e^x}{\sqrt{1-e^{2x}}}$ (multiplying inside the root by $e^{2x}/e^{2x}$ and pulling $e^x$ out front); putting $e^x=t$ (so $e^x\,dx=dt$) gives a Formula-4 shape, $\int\dfrac{dt}{\sqrt{1-t^2}}=\sin^{-1}(e^x)+c$. …