4.8: Integrals Involving Arctrig Functions
- Page ID
- 527
Recall that
\[\dfrac{d}{dx} \tan^{-1} x = \dfrac{1}{1+x^2},\nonumber \]
\[\dfrac{d}{dx} \sin^{-1} x = \dfrac{1}{\sqrt{1-x^2}},\nonumber \]
\[\dfrac{d}{dx} \sec^{-1} x = \dfrac{1}{x\sqrt{x^2-1}}.\nonumber \]
These three formulas immediately imply to integration:
\[ \int \dfrac{1}{1+x^2} dx = \tan^{-1} x + C,\nonumber \]
\[ \int \dfrac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x + C,\nonumber \]
\[ \int \dfrac{1}{x\sqrt{x^2-1}} dx = \sec^{-1} x + C.\nonumber \]
Example 1
Solve \( \int \dfrac{dx}{4+x^2}\).
Solution
\[ \int \dfrac{dx}{4+x^2} = \dfrac{1}{4} \int \dfrac{dx}{1+\left( \dfrac{x}{2} \right)^2}.\nonumber \]
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 \(x = 2u-3\).
\[\begin{align*} \int \dfrac{(2u-3)\, du}{u^2+1} &= \int \dfrac{2u\,du}{u^2+1}-3\int\dfrac{du}{u^2+1} \\ &= \ln \left| u^2+1 \right| -3\tan^{-1} u+C \\ &= \ln \left(\dfrac{ \left(\dfrac{x+3}{2} \right)^2}{4} +1\right) - 3 \tan^{-1}\dfrac{x+3}{2} + C \end{align*}.\nonumber \]
Contributors and Attributions
- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.