Trigonometric and Inverse Substitutions

Definition Let $f$ be continuous on an open interval $I$, and let $x = g(\theta)$ where $g$ is differentiable on an open interval $J$, $g'(\theta) \neq 0$ for every $\theta$ in $J$, and $g$ maps $J$ onto $I$. Then $g$ is one-to-one, its inverse $g^{-1}$ is differentiable on $I$, and

The equation means that if $G$ is an antiderivative of $f\big(g(\theta)\big) \, g'(\theta)$ on $J$, then $G\big(g^{-1}(x)\big)$ is an antiderivative of $f$ on $I$, so the answer can be written back in terms of $x$.

Definition For a constant $a > 0$,

Here $\arcsin$ takes values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $\arctan$ in $(-\frac{\pi}{2}, \frac{\pi}{2})$. For $\lvert y \rvert \ge 1$, $\operatorname{arcsec} y$ is the angle in $[0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]$ whose secant is $y$, so $\operatorname{arcsec} y = \arccos \frac{1}{y}$. With this convention

and that is why the third formula needs $\lvert x \rvert$. Some books use the range $[0, \frac{\pi}{2}) \cup [\pi, \frac{3\pi}{2})$ instead, and their version of the formula looks different for negative $x$.

Definition For an integrand containing $\sqrt{a^2 - x^2}$ with $a > 0$ and $-a \le x \le a$, put

Then $a^2 - x^2 = a^2 \cos^2 \theta$, and since $\cos \theta \ge 0$ on this interval, $\sqrt{a^2 - x^2} = a \cos \theta$.

Definition For an integrand containing $\sqrt{a^2 + x^2}$ or $a^2 + x^2$ with $a > 0$, put

Then $a^2 + x^2 = a^2 \sec^2 \theta$, and since $\sec \theta > 0$ on this interval, $\sqrt{a^2 + x^2} = a \sec \theta$.

Definition For an integrand containing $\sqrt{x^2 - a^2}$ with $a > 0$ and $\lvert x \rvert \ge a$, put

with $0 \le \theta < \frac{\pi}{2}$ when $x \ge a$ and $\frac{\pi}{2} < \theta \le \pi$ when $x \le -a$. Then $x^2 - a^2 = a^2 \tan^2 \theta$ and $\sqrt{x^2 - a^2} = a \lvert \tan \theta \rvert$, which equals $a \tan \theta$ when $x \ge a$ and $-a \tan \theta$ when $x \le -a$.

Definition A trigonometric substitution $x = g(\theta)$ must use an interval of $\theta$ on which $g$ is one-to-one and which $g$ maps onto the whole interval of $x$ you are working on. The standard choices come from the ranges of the inverse functions. They are $[-\frac{\pi}{2}, \frac{\pi}{2}]$ for $x = a \sin \theta$ and $(-\frac{\pi}{2}, \frac{\pi}{2})$ for $x = a \tan \theta$. For $x = a \sec \theta$ the integrand lives on two separate intervals, $x \ge a$ and $x \le -a$, so you work on one at a time, with $[0, \frac{\pi}{2})$ for $x \ge a$ and $(\frac{\pi}{2}, \pi]$ for $x \le -a$. On these ranges $\theta = \arcsin \frac{x}{a}$, $\theta = \arctan \frac{x}{a}$ or $\theta = \operatorname{arcsec} \frac{x}{a}$ exactly.

Definition For every real number $w$, $\sqrt{w^2} = \lvert w \rvert$. In a trigonometric substitution this means $\sqrt{a^2 \cos^2 \theta} = a \lvert \cos \theta \rvert$, $\sqrt{a^2 \sec^2 \theta} = a \lvert \sec \theta \rvert$ and $\sqrt{a^2 \tan^2 \theta} = a \lvert \tan \theta \rvert$. The absolute value can be dropped only when the branch guarantees the sign.

Definition To write a result in $\theta$ back in terms of $x$, use a right triangle with an acute angle $\theta$ whose sides match the substitution. For $x = a \sin \theta$ the opposite side is $x$, the hypotenuse $a$ and the adjacent side $\sqrt{a^2 - x^2}$. For $x = a \tan \theta$ the opposite side is $x$, the adjacent side $a$ and the hypotenuse $\sqrt{a^2 + x^2}$. For $x = a \sec \theta$ the hypotenuse is $x$, the adjacent side $a$ and the opposite side $\sqrt{x^2 - a^2}$. The triangle is drawn for $0 < \theta < \frac{\pi}{2}$, but the formulas it gives for the sine and tangent substitutions hold on the whole branch, since they follow from the identities and the sign of $\cos \theta$ or $\sec \theta$ there. For the secant substitution with $x \le -a$, check signs using the previous section.