Indefinite Integrals of Inverse Trigonometric Functions

Theorem 1: The following functions have the following indefinite integrals: a) If $f(x) = \sin ^{-1} x$, then $\int \sin ^{-1} x \: dx = x \sin ^{-1} x + \sqrt{1 - x^2} + C$. b) If $f(x) = \cos ^{-1} x$, then $\int \cos ^{-1} x \: dx = x\cos ^{-1} x - \sqrt{1 - x^2} + C$. c) If $f(x) = \tan ^{-1} x$, then $\int \tan ^{-1} x \: dx = x \tan ^{-1} x - \frac{1}{2} \ln \mid x^2 + 1 \mid + C$.

We will only prove a), however, the other integrals follow the same derivation. The reader is advised to mimic the proof of part a) to verify parts b) and c).

• Proof of a): Let $u = \sin ^{-1} x$ so that $dv = dx$. Therefore, $du = \frac{1}{\sqrt{1 - x^2}} \: dx$ and $v = x$. Using the integration by parts formula we obtain that:

(1)

\begin{align} \int \sin ^{-1} x \: dx = x \sin^{-1} x - \int \frac{x}{\sqrt{1 - x^2}} \: dx \end{align}

• Now we will use substitution. Let $u = 1 - x^2$ so that $du = -2x \: dx$. Furthermore, $-\frac{du}{2} = x dx$. Making this substitution we get that:

(2)

\begin{align} \int \sin ^{-1} x \: dx = x \sin^{-1} x - \int \frac{x}{\sqrt{1 - x^2}} \: dx \\ \int \sin ^{-1} x \: dx = x \sin^{-1} x - \int -\frac{1}{2} \frac{1}{\sqrt{u}} \: dx \\ \int \sin ^{-1} x \: dx = x \sin^{-1} x + \frac{1}{2} \int u^{-1/2} \: du \\ \int \sin ^{-1} x \: dx = x \sin^{-1} x + (1/2) \frac{u^{1/2}}{(1/2)} \\ \int \sin ^{-1} x \: dx = x \sin^{-1} x + u^{1/2} \\ \int \sin ^{-1} x \: dx = x \sin^{-1} x + \sqrt{1 - x^2} + C \quad \blacksquare \\ \end{align}