7. Trigonometric Substitutions

d. Hyperbolic Trig Substitutions (Optional)

The kind of substitutions we have done with the trig functions \(\tan\), \(\sin\) and \(\sec\), we can also do with the hyperbolic trig functions. The relevant identities are: \[\begin{aligned} \cosh^2\lambda-\sinh^2\lambda=1 \\ 1-\tanh^2\lambda=\mathrm{sech}^2\lambda \\ \coth^2\lambda-1=\mathrm{csch}^2\lambda \end{aligned}\] along with the derivative rules: \[\begin{aligned} \dfrac{d}{dx}\sinh x&=\cosh x \qquad &\dfrac{d}{dx}\cosh x&=\sinh x \\ \dfrac{d}{dx}\tanh x&=\,\mathrm{sech}^2x \qquad &\dfrac{d}{dx}\coth x&=-\,\mathrm{csch}^2x \\ \dfrac{d}{dx}\,\mathrm{sech}\,x&=-\,\mathrm{sech}\,x\tanh x \qquad &\dfrac{d}{dx}\,\mathrm{csch}\,x&=-\,\mathrm{csch}\,x\coth x \end{aligned}\]

Compute \(\displaystyle \int \dfrac{1}{\sqrt{9x^2+4}}\,dx\).

Let \(9x^2=4\sinh^2\lambda\) or \(x=\dfrac{2}{3}\sinh\lambda\). Then \(dx=\dfrac{2}{3}\cosh\lambda\,d\lambda\) and so \[\begin{aligned} \int \dfrac{1}{\sqrt{9x^2+4}}\,dx &=\int \dfrac{1}{\sqrt{4\sinh^2\lambda+4}}\dfrac{2}{3}\cosh\lambda\,d\lambda\\ &=\dfrac{1}{3}\int \dfrac{1}{\sqrt{\cosh^2\lambda}}\cosh\lambda\,d\lambda =\dfrac{1}{3}\int \,d\lambda \\ &=\dfrac{1}{3}\lambda+C \end{aligned}\]

To substitute back, we know \(\lambda=\mathrm{arcsinh}\left(\dfrac{3x}{2}\right)\). So: \[ \int \dfrac{1}{\sqrt{9x^2+4}}\,dx= \dfrac{1}{3}\mathrm{arcsinh}\left(\dfrac{3x}{2}\right) + C \]

This is a fine answer, but it may be nice to re-express it in terms of logs using the formula \(\mathrm{arcsinh}\,z=\ln(z+\sqrt{z^2+1})\): \[\begin{aligned} \int \dfrac{1}{\sqrt{9x^2+4}}\,dx &=\dfrac{1}{3}\ln\left(\dfrac{3x}{2}+\dfrac{\sqrt{9x^2+4}}{2}\right)+C \\ &=\dfrac{1}{3}\ln\left(3x+\sqrt{9x^2+4}\right)-\dfrac{1}{3}\ln2+C \end{aligned}\] This is the same answer found using a normal \(\tan\) substitution but the computation was shorter.

We check by differentiating. Let \(f(x)=\dfrac{1}{3}\mathrm{arcsinh}\left(\dfrac{3x}{2}\right)\). The derivative of \(\mathrm{arcsinh}\,x\) is \(\dfrac{1}{\sqrt{1+x^2}}\). Thus, \[\begin{aligned} f'(x) &=\dfrac{1}{3}\dfrac{1}{\sqrt{1+\dfrac{9x^2}{4}}}\dfrac{3}{2} \\ &=\dfrac{1}{2}\dfrac{2}{\sqrt{9x^2+4}} =\dfrac{1}{\sqrt{9x^2+4}} \end{aligned}\] which is the integrand we started with. The check was easier too.