4.8: Integrals Involving Arctrig Functions

Recall that

\[\dfrac{d}{dx} \tan^{-1} x = \dfrac{1}{1+x^2},\]

\[\dfrac{d}{dx} \sin^{-1} x = \dfrac{1}{\sqrt{1-x^2}},\]

\[\dfrac{d}{dx} \sec^{-1} x = \dfrac{1}{x\sqrt{x^2-1}}.\]

These three formulas immediately imply to integration:

\[ \int \dfrac{1}{1+x^2} dx = \tan^{-1} x + C,\]

\[ \int \dfrac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x + C,\]

\[ \int \dfrac{1}{x\sqrt{x^2-1}} dx = \sec^{-1} x + C.\]

Example 1

Solve \( \int \dfrac{dx}{4+x^2}\).

Solution

\[ \int \dfrac{dx}{4+x^2} = \dfrac{1}{4} \int \dfrac{dx}{1+\left( \dfrac{x}{2} \right)^2}.\]

Let \(u= \dfrac{x}{2}\) so \(du=\dfrac{1}{2}dx\)

and the integral becomes

\[\begin{align} \dfrac{1}{2} \int \dfrac{du}{1+u^2} = \dfrac{1}{2} \tan^{-1} u + C &= \dfrac{1}{2} \tan^{-1} \left( \dfrac{x}{2} \right) + C. \end{align}\]

Example 2

Solve \( \int \dfrac{dx}{x \sqrt{x^2-4}} \).

Solution

\[\begin{align} \int \dfrac{dx}{x \sqrt{x^2-4}} &= \int \dfrac{x\,dx}{x^2 \sqrt{4 (x^4/4 -1)}} \\ &= \dfrac{1}{4} \int \dfrac{x\,dx}{(x^2/2)\sqrt{(x^2/2)^2-1}} \end{align}\]

which becomes

\[\begin{align} \dfrac{1}{4} \int \dfrac{du}{u\sqrt{u^2-1}} &= \dfrac{1}{4} \sec^{-1} u + C \\ &= \dfrac{1}{4} \sec^{-1} \left(\dfrac{x^2}{2}\right) + C. \end{align}\]

Example 3

Solve \(\int \dfrac{2x\, dx}{x^2+6x+13}\).

Solution

\[\begin{align} \int \dfrac{2x\, dx}{x^2+6x+13} &= \int \dfrac{2x\, dx}{(x+3)^2 + 4} \\ &=\dfrac{1}{2} \int \dfrac{x\, dx}{ \left(\dfrac{x+3}{2} \right)^2+1} \end{align}\]

let \(u= \dfrac{x+3}{2}\) and \(du = \dfrac{1}{2} dx\) so \(x = 2u-3\).

\[\begin{align} \int \dfrac{(2u-3)\, du}{u^2+1} &= \int \dfrac{2u\,du}{u^2+1}-3\int\dfrac{du}{u^2+1} \\ &= \ln \left| u^2+1 \right| -3\tan^{-1} u+C \\ &= \ln \left(\dfrac{ \left(\dfrac{x+3}{2} \right)^2}{4} +1\right) - 3 \tan^{-1}\dfrac{x+3}{2} + C \end{align}.\]

Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.