Integral using multiple methods

Question

I'm teaching a Calculus II class and we are covering integration techniques. We've covered $u$-substitution, integration by parts, trig integrals, trigonometric substitutions, partial fractions and integration involving hyperbolic functions. I want to give my students some integrals to do that can be done using as many of these methods as possible. So far, I've come up with $\int\frac{x}{x^2-1}dx$ which can be done using partial fractions, $u$-substitution, trigonometric substitution, and a hyperbolic cosine substitution. I was wondering if anyone had other examples like this. Thanks.

Answer 1

Perhaps these will be useful: I solve three integrals below by both trigonometric and hyperbolic substitutions. You can see the mirroring of calculational difficulty. I like to teach both.

1. [trig]We wish to calculate $\int \sqrt{x^2-9} \ dx$. The squareroot is trouble. One sneaky way to eliminate it is to let $x = 3\sec(\theta)$ thus $x^2-9 = 9\sec^2(\theta) - 9 = 9\tan^2(\theta)$ and $dx = 3\sec(\theta)\tan(\theta)d\theta$. Hence, \begin{align} \notag \int \sqrt{x^2-9} dx &= \int \sqrt{9\tan^2(\theta)} 3\sec(\theta)\tan(\theta)d\theta \\ \notag &= \int 9\sec(\theta)\tan^2(\theta)d\theta \\ \notag &= \int 9\sec(\theta)(\sec^2(\theta)-1)d\theta \\ \notag &= 9\int \sec^3(\theta)d\theta - 9\int \sec(\theta)d\theta\\ \notag &= \frac{9}{2}\sec(\theta)\tan(\theta) + \frac{9}{2}\ln|\sec(\theta)+\tan(\theta)| - 9\ln|\sec(\theta)+\tan(\theta)| +c \\ \notag &= \frac{9}{2}\sec(\theta)\tan(\theta) - \frac{9}{2}\ln|\sec(\theta)+\tan(\theta)|+c\\ \notag \end{align} We can simplify this answer nicely if we think about the substitution in terms of a triangle. Note that if $x = 3\sec(\theta)$ then $\sec(\theta) = \frac{x}{3} = \frac{ hyp}{adj}$ hence $opp = \sqrt{x^2-9}$ and so $\tan(\theta) = \frac{\sqrt{x^2-9}}{3}$. We find, $$ \boxed{\int \sqrt{x^2-9} \ dx = \frac{1}{2}x\sqrt{x^2-9}- \frac{9}{2}\ln \bigg|\frac{x}{3}+\frac{\sqrt{x^2-9}}{3}\bigg|+c.} $$

2. [hyperbolic]The fundamental identity for hyperbolic trigonometry is $\cosh^2(\phi)-\sinh^2(\phi)=1$. We can rearrange this to give $\cosh^2(\phi)-1 = \sinh^2(\phi)$. This suggests we might be able to work the previous example by making a $x=3\cosh(\phi)$ substitution. If $x=3\cosh(\phi)$ then $dx = 3\sinh(\phi)d\phi$ and $x^2-9 = 9\sinh^2(\phi)$. Hence, $$ \int \sqrt{x^2-9} \ dx = \int 9\sinh^2(\phi) d\phi. $$ I need an identity for $\sinh^2(\phi)$, let's derive it from scratch, $$ \sinh^2(\phi) = \biggl[\frac{1}{2}\bigl(e^{\phi}-e^{-\phi} \bigr)\biggr]^2 = \frac{1}{4}\biggl[e^{2\phi}-2+e^{-2\phi} \biggr] = \frac{1}{2}\cosh(2\phi)-\frac{1}{2} $$ Returning to our integration with this new-found insight, $$ \int \sqrt{x^2-9} \ dx = \int \biggl( \frac{9}{2}\cosh(2\phi)-\frac{9}{2} \biggr) d\phi = \frac{9}{4}\sinh(2\phi)-\frac{9\phi}{2}+c. $$ Note that $2\sinh(\phi)\cosh(\phi) = \frac{1}{2}(e^{\phi}-e^{-\phi})(e^{\phi}+e^{-\phi}) = \frac{1}{2}(e^{2\phi}-e^{-2\phi}) = \sinh(2\phi)$. Furthermore, we began by supposing that $x=3\cosh(\phi)$ hence $\cosh(\phi) = \frac{x}{3}$ and $\sinh(\phi) = \sqrt{x^2-9}$. We find that $\frac{9}{4}\sinh(2\phi) = \frac{9}{2}\frac{x}{3}\frac{\sqrt{x^2-9}}{3} = \frac{1}{2}x\sqrt{x^2-9}$. On the other hand, in our current formalism we are led to write $\frac{9\phi}{2} = \frac{9}{2}\cosh^{-1}(\frac{x}{3})$. Thus, $$ \boxed{\int \sqrt{x^2-9} \ dx = \frac{1}{2}x\sqrt{x^2-9}-\frac{9}{2}\cosh^{-1}\biggl(\frac{x}{3}\biggr)+c.} $$

3. [trig]We wish to calculate $\int \sqrt{4x^2+9} \ dx$. The squareroot is trouble. One sneaky way to eliminate it is to let $2x = 3\tan(\theta)$ thus $4x^2+9 = 9\tan^2(\theta) + 9 = 9\sec^2(\theta)$ and $2dx = 3\sec^2(\theta)d\theta$. Hence, \begin{align} \notag \int \sqrt{4x^2+9} \ dx &= \int \sqrt{9\sec^2(\theta)} \ \frac{3\sec^2(\theta)d\theta}{2} \\ \notag &= \frac{9}{2}\int \sec^3(\theta)d\theta \\ \notag &= \frac{9}{4}\sec(\theta)\tan(\theta) + \frac{9}{4}\ln|\sec(\theta)+\tan(\theta)| +c \\ \notag \end{align} We can simplify this answer nicely if we think about the substitution in terms of a triangle. Note that if $2x = 3\tan(\theta)$ then $\tan(\theta) = \frac{2x}{3} = \frac{ opp}{adj}$ hence $hyp = \sqrt{4x^2+9}$ and so $\sec(\theta) = \frac{\sqrt{4x^2+9}}{3}$. We find, $$ \int \sqrt{4x^2+9} \ dx = \frac{1}{2}x\sqrt{4x^2+9}+ \frac{9}{4}\ln\bigg|\frac{\sqrt{4x^2+9}}{3}+\frac{2x}{3}\bigg| +c. $$ Notice, we can simplify this answer by factoring out a $1/3$ in the natural log argument, $$ \boxed{\int \sqrt{4x^2+9} \ dx = \frac{1}{2}x\sqrt{4x^2+9}+ \frac{9}{4}\ln\big|\sqrt{4x^2+9}+2x \big| +c.} $$

4. [hyperbolic] Another way to calculate $\int \sqrt{4x^2+9} \ dx$ is to make a hyperbolic substitution. Observe that $\cosh^2(\phi)-\sinh^2(\phi)=1$ gives $\cosh^2(\phi) = \sinh^2(\phi)+1$. This suggests we can make a $2x = 3\sinh(\phi)$ substitution. If $2x = 3\sinh(\phi)$ then $4x^2+9 = 9\sinh^2(\phi)+9 = 9 \cosh^2(\phi)$ and $2dx = 3\cosh(\phi)d\phi$. \begin{align} \notag \int \sqrt{4x^2+9} \ dx &= \int \sqrt{9 \cosh^2(\phi)} \ \frac{3\cosh(\phi)d\phi}{2} \\ \notag &= \frac{9}{2}\int \cosh^2(\phi)d\phi \\ \notag \end{align} I need an identity for $\cosh^2(\phi)$, let's derive it from scratch, $$ \cosh^2(\phi) = \biggl[\frac{1}{2}\bigl(e^{\phi}+e^{-\phi} \bigr)\biggr]^2 = \frac{1}{4}\biggl[e^{2\phi}+2+e^{-2\phi} \biggr] = \frac{1}{2}\cosh(2\phi)+\frac{1}{2} $$ Returning to our integration with this new-found insight, \begin{align} \notag \int \sqrt{4x^2+9} \ dx &= \frac{9}{4}\int \bigl(\cosh(2\phi)+1\bigr)d\phi \\ \notag &= \frac{9}{8}\sinh(2\phi)+\frac{9}{4}\phi + c. \\ \notag &= \frac{9}{4}\frac{2x}{3}\frac{\sqrt{4x^2+9}}{3}+\frac{9}{4}\phi + c. \\ \notag &= \boxed{ \frac{1}{2}x\sqrt{4x^2+9}+\frac{9}{4}\sinh^{-1} \biggl( \frac{2x}{3} \biggr) + c.} \end{align} I used the identity $\sinh(2\phi) = 2\sinh(\phi)\cosh(\phi)$ to help simplify the answer.

5. [trig] To begin I simplify the quadratic by completing the square: \begin{align} \notag \int \frac{dx}{x^2+6x+5} &= \int \frac{dx}{(x+3)^2-4}. \\ \notag \end{align} Remember that $\tan^2(\theta)=\sec^2(\theta)-1$ so this makes me think a $(x+3)^2 = 4\sec^2(\theta)$ substitution will simplify this problem. If $x+3 = 2\sec(\theta)$ then $(x+3)^2-4 = 4\sec^2(\theta)-4 = 4\tan^2(\theta)$ and $dx = 2\sec(\theta)\tan(\theta)d\theta$. Thus, \begin{align} \notag \int \frac{dx}{x^2+6x+5} &= \int \frac{2\sec(\theta)\tan(\theta)d\theta}{4\tan^2(\theta)} \\ \notag &= \int \frac{\sec(\theta)d\theta}{2\tan(\theta)} \\ \notag &= \frac{1}{2}\int \csc(\theta) d\theta \\ \notag &= \frac{-1}{2}\int \frac{du}{u} \qquad \text{let $u = \csc(\theta) + \cot(\theta)$ hence $-du/u = \csc(\theta) d\theta$} \\ \notag &= \frac{-1}{2} \ln|\csc(\theta) + \cot(\theta)| + c \\ \notag \end{align} We let $x+3 = 2\sec(\theta)$ thus $\sec(\theta) = \frac{x+3}{2} = \frac{hyp}{adj}$. The other side of the substitution triangle is found by the pythagorean theorem; $opp = \sqrt{(x+3)^2-4}$. Observe that $\cot(\theta) = \frac{adj}{opp} = \frac{2}{ \sqrt{(x+3)^2-4}}$ and $\csc(\theta) = \frac{hyp}{opp} = \frac{x+3}{ \sqrt{(x+3)^2-4}}$. Therefore, \begin{align} \notag \int \frac{dx}{x^2+6x+5} &= \boxed{ \frac{-1}{2} \ln\bigg|\frac{x+3}{ \sqrt{(x+3)^2-4}} + \frac{2}{ \sqrt{(x+3)^2-4}}\bigg| + c. } \end{align}

6. [hyperbolic] If $\cosh^2(\phi)-\sinh^2(\phi)=1$ then dividing by $\sinh^2(\phi)$ reveals the hyperbolic cotangent/cosecant identity: $\coth^2(\phi)-1 = csch^2(\phi)$ where by definition $\coth(\phi) = \frac{\cosh(\phi)}{\sinh(\phi)}$ and $csch(\phi) = \frac{1}{\sinh(\phi)}$. In view of the identity above, the integral $\int \frac{dx}{x^2+6x+5} = \int \frac{dx}{(x+3)^2-4}$ is likely simplified by an $(x+3)^2 = 4\coth^2(\phi)$ substitution. If $x+3 = 2\coth(\phi)$ then $(x+3)^2-4 = 4\coth^2(\phi)-4 = 4csch^2(\phi)$ and \footnote{just use the quotient rule; $\coth(\phi)' = \frac{\cosh(\phi)}{\sinh(\phi)}' = \frac{\sinh^2(\phi)-\cosh^2(\phi)}{\sinh^2(\phi)} = \frac{-1}{\sinh^2(\phi)} = -csch^2(\phi)$} $dx = -2csch^2(\phi)d\phi$. \begin{align} \notag \int \frac{dx}{x^2+6x+5} &= \int \frac{dx}{(x+3)^2-4}. \\ \notag &= \int \frac{-2csch^2(\phi)d\phi}{4csch^2(\phi)}. \\ \notag &= \frac{-1}{2}\phi+c. \\ \notag &= \boxed{\frac{-1}{2}\coth^{-1}\biggl(\frac{x+3}{2} \biggr)+c.} \\ \notag \end{align}