Reading the inverse trig derivatives backward gives two new antiderivatives. They answer integrals that look like log integrals but are not.
Key idea
$$\int\frac{1}{1 + x^2}\,dx = \arctan x + C, \qquad \int\frac{1}{\sqrt{1 - x^2}}\,dx = \arcsin x + C,$$ and, for $a > 0$, $$\int\frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\arctan\frac{x}{a} + C, \qquad \int\frac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\frac{x}{a} + C.$$
Why it works
The first pair is the derivative table read backward. For the versions with $a$, factor $a^2$ out of the bottom and substitute $u = \frac{x}{a}$: $\frac{1}{a^2 + x^2} = \frac{1}{a^2}\cdot\frac{1}{1 + (x/a)^2}$, and $dx = a\,du$, so the integral becomes $\frac{1}{a}\int\frac{du}{1 + u^2} = \frac{1}{a}\arctan u$. The arcsine case works the same way, and the $a$s cancel completely.
Worked example: With a constant
Find $\displaystyle\int\frac{dx}{9 + x^2}$.
- $a^2 = 9$, so $a = 3$.
- $\frac{1}{3}\arctan\frac{x}{3} + C$. Check: $\frac{d}{dx}\frac{1}{3}\arctan\frac{x}{3} = \frac{1}{3}\cdot\frac{1/3}{1 + x^2/9} = \frac{1}{9 + x^2}$.
Answer: $\frac{1}{3}\arctan\frac{x}{3} + C$
Worked example: A definite arcsine
Evaluate $\displaystyle\int_0^1\frac{dx}{\sqrt{4 - x^2}}$.
- $a = 2$: antiderivative $\arcsin\frac{x}{2}$.
- $\arcsin\frac{1}{2} - \arcsin 0 = \frac{\pi}{6} - 0$.
Answer: $\frac{\pi}{6}$
Worked example: Complete the square
Find $\displaystyle\int\frac{dx}{x^2 + 4x + 13}$.
- Complete the square: $x^2 + 4x + 13 = (x + 2)^2 + 9$.
- $u = x + 2$, $du = dx$: $\int\frac{du}{u^2 + 9} = \frac{1}{3}\arctan\frac{u}{3} + C$.
Answer: $\frac{1}{3}\arctan\frac{x + 2}{3} + C$
| Integrand | Method | Antiderivative |
|---|---|---|
| $\frac{x}{1 + x^2}$ | $u = 1 + x^2$ | $\frac{1}{2}\ln(1 + x^2)$ |
| $\frac{1}{1 + x^2}$ | standard form | $\arctan x$ |
| $\frac{x}{\sqrt{1 - x^2}}$ | $u = 1 - x^2$ | $-\sqrt{1 - x^2}$ |
| $\frac{1}{\sqrt{1 - x^2}}$ | standard form | $\arcsin x$ |
Integrals like $\int\frac{dx}{1 - x^2}$ (no root, a minus sign) need yet another method, partial fractions, which comes in Calculus 2. The shape of the denominator tells you which tool to reach for.
Common mistake
Writing $\ln(1 + x^2)$ for $\int\frac{1}{1 + x^2}\,dx$. The log answer needs the derivative of the bottom on top: $\int\frac{2x}{1 + x^2}\,dx = \ln(1 + x^2) + C$. With no $x$ on top, the answer is an arctangent. Check by differentiating: $\frac{d}{dx}\ln(1 + x^2) = \frac{2x}{1 + x^2}$, not $\frac{1}{1 + x^2}$.