How to Use the Reverse Chain Rule with Inverse Trig Functions

Recognize the derivative pair

Some integrals contain an inverse trigonometric function together with its derivative. These are often simpler than they look because substitution turns them into a power integral.

Three common inverse-trig derivatives are

$$\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}},$$ $$\frac{d}{dx}\arctan x=\frac{1}{1+x^2},$$

and

$$\frac{d}{dx}\operatorname{arcsec}x=\frac{1}{|x|\sqrt{x^2-1}}.$$

After a chain-rule substitution, an integral of the form

$$\int [g(x)]^n g'(x)\,dx$$

becomes $\int u^n\,du$, where $u=g(x)$.

Example with arctangent

Consider

$$\int \frac{\arctan(3x)}{1+9x^2}\,dx.$$

Let $u=\arctan(3x)$. Then

$$du=\frac{3}{1+9x^2}\,dx,$$

so $dx/(1+9x^2)=du/3$. Therefore,

$$\int \frac{\arctan(3x)}{1+9x^2}\,dx =\frac13\int u\,du =\frac16u^2+C =\frac16[\arctan(3x)]^2+C.$$

Differentiate the result to verify both the outer power factor and the inner chain-rule factor.

Do not confuse two problem families

• An integral such as $\int dx/(1+x^2)$ has an inverse-trig function as its result.
• An integral such as the example above already contains an inverse-trig function and uses its derivative as the substitution factor.

Checking which pattern is present determines the correct first step.