Section 1.17 Integration strategies
Having introduced a wealth of new derivative/integral formulas and rules, we now take a moment to give an overview of our integration techniques, and apply them in combination with some algebraic methods to solving integrals in the wild.
Theorem 1.17.1. Idefinite integral compendium.
We collect here our various derivative/antiderivative formulas.
\begin{align*}
\frac{d}{dx} x^r=rx^{r-1}\amp \iff \int x^r \, dx=\frac{1}{r+1}x^{r+1}+C, r\ne -1\\
\frac{d}{dx} \sin x=\cos x\amp \iff \int\cos x \, dx=\sin x+C\\
\frac{d}{dx}\cos x=-\sin x \amp \iff \int \sin x \, dx=-\cos x+C\\
\frac{d}{dx}\tan x=\sec^2x \amp \iff \int \sec^2 x \, dx=\tan x+C\\
\frac{d}{dx}\cot x=-\csc^2x \amp \iff \int \csc ^2x\, dx=-\cot x+C\\
\frac{d}{dx}\sec x=\sec x\tan x \amp \iff \int \sec x\tan x \, dx=\sec x+C\\
\frac{d}{dx}\csc x=-\csc x\cot x \amp \iff \int\csc x\cot x \, dx=-\csc x+C\\
\frac{d}{dx}\ln\vert x\vert=\frac{1}{x} \amp \iff \int\frac{1}{x} \, dx=\ln\vert x\vert+C\\
\frac{d}{dx}\ln\vert \cos x \vert=-\tan x \amp \iff \int\tan x \, dx=-\ln\vert \cos x \vert+C=\ln\vert\sec x\vert+C\\
\frac{d}{dx}\ln\vert \sin x\vert=\cot x \amp \iff \int\cot x \, dx=\ln\vert \sin x\vert+C\\
\frac{d}{dx}\ln\vert \sec x+\tan x\vert=\sec x \amp \iff \int\sec x \, dx=\ln\vert \sec x+\tan x\vert+C\\
\frac{d}{dx}\ln\vert \csc x+\cot x\vert=-\csc x \amp \iff \int\csc x \, dx=-\ln\vert \csc x+\cot x\vert+C\\
\frac{d}{dx}e^x=e^x \amp \iff \int e^x \, dx=e^x+C\\
\frac{d}{dx}a^x=(\ln a)a^x \amp \iff \int a^x \, dx=\frac{1}{\ln a}a^x+C\\
\frac{d}{dx}\log_a \vert x\vert =\frac{1}{(\ln a)\, x} \amp \iff \int \frac{1}{(\ln a)\, x} \, dx=\log_a\vert x\vert+C\\
\frac{d}{dx} \arcsin x=\frac{1}{\sqrt{1-x^2}}\amp \iff \int \frac{1}{\sqrt{1-x^2}}\, dx=\arcsin x+C\\
\frac{d}{dx} \arccos x=-\frac{1}{\sqrt{1-x^2}}\amp \iff \int \frac{1}{\sqrt{1-x^2}}\, dx=-\arccos x+C\\
\frac{d}{dx} \arctan x=\frac{1}{1+x^2}\amp \iff \int \frac{1}{1+x^2}\, dx=\arctan x+C \text{.}
\end{align*}
Each of the integral computations below will combine various integral formulas, substitution, and an algebraic method.
Example 1.17.2. Vertex form.
Compute \(\displaystyle \int \frac{1}{x^2-6x+18}\, dx\text{.}\)
Solution.
We have
\begin{align*}
\int \frac{1}{x^2-6x+18}\, dx \amp = \int \frac{1}{(x-3)^2+9}\, dx \\
\amp = \frac{1}{9}\int\frac{1}{((x-3)/3)^2+1}\\
\amp =\frac{1}{3}\int \frac{1}{u^2+1} \amp (u=(x-3)/3, du=\frac{1}{3}dx)\\
\amp = \arctan((x-3)/3)+C\text{.}
\end{align*}
Example 1.17.3. Polynomial division.
Compute \(\displaystyle \int \frac{4x^3+3x+1}{4x^2+1}\, dx\text{.}\)
Hint.
Use polynomial division with remainder.
Solution.
Using polynomial division with remainder, we find that
\begin{equation*}
\frac{4x^3+3x+1}{4x^2+1}=x+\frac{2x+1}{4x^2+1}\text{,}
\end{equation*}
and thus
\begin{align*}
\int \frac{4x^3+3x+1}{4x^2+1}\, dx \amp = \int x+\frac{2x+1}{4x^2+1}\, dx \\
\amp = \int x+\frac{2x}{4x^2+1}+\frac{1}{4x^2+1}\, dx\\
\amp = \frac{1}{2}x^2+\frac{1}{4}\ln\abs{4x^2+1}+\frac{1}{2}\arctan(2x)+C\text{.}
\end{align*}
Example 1.17.4. Exponential substitution.
Compute \(\displaystyle\int \frac{1}{e^x+e^{-x}}\, dx\text{.}\)
Solution.
We have
\begin{align*}
\int \frac{1}{e^x+e^{-x}}\, dx \amp = \int \frac{1}{u+1/u}\cdot \frac{1}{u}\, du \amp (u=e^x, du=e^x dx\implies dx=\frac{1}{u}du) \\
\amp = \int \frac{1}{u^2+1}\, du\\
\amp = \arctan(e^x)+C\text{.}
\end{align*}
Example 1.17.5. Exponential substitution (again).
Compute \(\displaystyle \int \frac{3}{\sqrt{e^{2x}-2}}\, dx\text{.}\)
Solution.
We have
\begin{align*}
\int \frac{3}{\sqrt{e^{2x}-2}}\, dx \amp = \int \frac{3}{e^x\sqrt{1-2e^{-2x}}} \amp
(e^{2x}=(e^x)^2, 1/e^{2x}=e^{-2x})\\
\amp = -\frac{3}{\sqrt{2}}\int \frac{1}{\sqrt{1-u^2}}\, du \amp (u=\sqrt{2}e^{-x}, du=-\sqrt{2}e^{-x}dx\implies dx=-\frac{1}{\sqrt{2}u}du)\\
\amp = -\frac{3}{\sqrt{2}}\arcsin(\sqrt{2}e^{-x})+C\text{.}
\end{align*}
Example 1.17.6. Trig identity.
Compute \(\displaystyle \int_0^{\pi/3}\sin^2(2x)\cos(3x) \, dx\text{.}\)
Hint.
Make use of some of the following product-to-sum identities.
\begin{align*}
\cos\theta\cos\phi\amp =\frac{\cos(\theta-\phi)+\cos(\theta+\phi)}{2}\\
\sin\theta\sin\phi\amp =\frac{\cos(\theta-\phi)-\cos(\theta+\phi)}{2}\\
\sin\theta\cos\phi\amp = \frac{\sin(\theta-\phi)+\sin(\theta+\phi)}{2}\\
\cos^2\theta\amp =\frac{1+\cos2\theta}{2}\\
\sin^2\theta\amp =\frac{1-\cos2\theta}{2}
\end{align*}
Solution.
We have
\begin{align*}
\int_0^{\pi/3}\sin^2(2x)\cos(3x) \, dx \amp = \int_0^{\pi/3}\frac{1}{2}(1-\cos(4x))\cos(3x) \, dx \\
\amp = \frac{1}{2}\int_0^{\pi/3}\cos(3x)-\cos(4x)\cos(3x)\, dx\\
\amp =\frac{1}{6}\sin(3x)\vert_0^{\pi/3}-\frac{1}{2}\int\frac{1}{2}(\cos x+\cos(7x))\, dx \\
\amp = 0-\left( \frac{1}{4}\sin x+\frac{1}{28}\sin 7x\right)\Bigr\vert_0^{\pi/3}\\
\amp = -\left( \frac{\sqrt{3}}{8}+\frac{\sqrt{3}}{56}\right)\\
\amp = -\frac{\sqrt{3}}{7}\text{.}
\end{align*}
