Calculus with Inverse Trig Functions

Starting point: what you already know

• Inverse trig functions and their domains and ranges (Trigonometric Functions unit).
• From Mathematics Advanced: the chain rule $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$, and $\frac{d}{dx}\sin x=\cos x$, $\frac{d}{dx}\tan x={\sec }^{2}x$.
• $\frac{dy}{dx}=\frac{1}{dx/dy}$, and the identities ${\sin }^2\theta +{\cos }^2\theta =1$ and $1+{\tan }^2\theta ={\sec }^2\theta$.
• Integration is the reverse of differentiation: $\int f^{\prime }(x)dx=f(x)+C$.

Finding the derivatives

Let $y={\sin }^{-1}x$, so $x=\sin y$ with $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$.

$$\frac{dx}{dy}=\cos y=\sqrt{1-{\sin }^{2}y}=\sqrt{1-x^2}$$

We took the positive root because $\cos y\ge 0$ in that range. Flip it over:

$$\frac{d}{dx}{\sin }^{-1}x=\frac{1}{\sqrt{1-x^2}}$$

The same idea gives the other two.

Integrals that give inverse trig

Reverse the derivatives:

Quick summary

• $\frac{d}{dx}{\sin }^{-1}x=\frac{1}{\sqrt{1-x^2}}$, $\frac{d}{dx}{\cos }^{-1}x=\frac{-1}{\sqrt{1-x^2}}$, $\frac{d}{dx}{\tan }^{-1}x=\frac{1}{1+x^2}$.
• Use the chain rule for ${\sin }^{-1}(f(x))$ and similar.
• $\int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\frac{x}{a}+C$ and $\int \frac{dx}{a^2+x^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C$.

Quiz