2.4 - Hyperbolic and inverse-form integrals

2.4 - Hyperbolic and inverse-form integrals

Recognising a derivative backwards can turn a complicated-looking integral into a short calculation. You will integrate hyperbolic functions, distinguish three inverse-function patterns and complete the square to reveal those patterns in quadratics.

Integrating hyperbolic functions

Why does integrating cosh⁡(3x)\cosh(3x) require division by 33, while differentiating it requires multiplication by 33? Integration must undo the whole chain rule: since (sinh⁡(3x))′=3cosh⁡(3x)(\sinh(3x))'=3\cosh(3x), an antiderivative is sinh⁡(3x)/3\sinh(3x)/3.

An antiderivative of ff is a function whose derivative is ff. Adding a constant does not change that derivative. Write +C+C for an indefinite integral, on an interval where the integrand is defined.

Basic hyperbolic integrals

∫sinh⁡x dx=cosh⁡x+C∫cosh⁡x dx=sinh⁡x+C∫sech⁡2x dx=tanh⁡x+C\begin{aligned} \int\sinh x\,\mathrm dx&=\cosh x+C \\ \int\cosh x\,\mathrm dx&=\sinh x+C \\ \int\operatorname{sech}^2x\,\mathrm dx&=\tanh x+C \end{aligned}

These three results are in the formula booklet, with arbitrary constants omitted there.

The reciprocal derivatives from the previous lesson give three more patterns:

∫cosech⁡2x dx=−coth⁡x+C∫sech⁡xtanh⁡x dx=−sech⁡x+C∫cosech⁡xcoth⁡x dx=−cosech⁡x+C\begin{aligned} \int\operatorname{cosech}^2x\,\mathrm dx&=-\coth x+C \\ \int\operatorname{sech}x\tanh x\,\mathrm dx&=-\operatorname{sech}x+C \\ \int\operatorname{cosech}x\coth x\,\mathrm dx&=-\operatorname{cosech}x+C \end{aligned}

These three integrals are not listed in the booklet: reverse the corresponding derivatives. Cosech and coth require nonzero arguments. For any of these patterns with argument kx+bkx+b, divide the antiderivative by kk, where k≠0k\ne0.

Worked example 1: undoing two inner scales

Question. Find ∫(6cosh⁡(3x)−4sech⁡2(2x)) dx\displaystyle\int\bigl(6\cosh(3x)-4\operatorname{sech}^2(2x)\bigr)\,\mathrm dx.

Plan. Reverse each basic derivative and divide by its own inner scale.

Linearity lets us integrate the terms separately:

∫6cosh⁡(3x) dx=63sinh⁡(3x)∫−4sech⁡2(2x) dx=−42tanh⁡(2x)\begin{aligned} \int6\cosh(3x)\,\mathrm dx&=\frac63\sinh(3x) \\ \int-4\operatorname{sech}^2(2x)\,\mathrm dx&=-\frac42\tanh(2x) \end{aligned}

Answer. 2sinh⁡(3x)−2tanh⁡(2x)+C2\sinh(3x)-2\tanh(2x)+C, for every real xx.

Check. Differentiating the answer gives 2(3cosh⁡(3x))−2(2sech⁡2(2x))2(3\cosh(3x))-2(2\operatorname{sech}^2(2x)), exactly the original integrand. At zero its derivative is 6−4=26-4=2, which also checks both coefficients.

Keep the coefficients and scales, but change to the sinh/cosech family. The derivative of coth has a minus sign, so integrating the second term now produces a plus sign.

Worked example 2: the corresponding reciprocal family

Question. Find ∫(6sinh⁡(3x)−4cosech⁡2(2x)) dx\displaystyle\int\bigl(6\sinh(3x)-4\operatorname{cosech}^2(2x)\bigr)\,\mathrm dx, stating the excluded value of xx.

Plan. Reverse the cosh and coth derivatives, retaining the sign in the coth rule.

∫6sinh⁡(3x) dx=2cosh⁡(3x)∫−4cosech⁡2(2x) dx=2coth⁡(2x)\begin{aligned} \int6\sinh(3x)\,\mathrm dx&=2\cosh(3x) \\ \int-4\operatorname{cosech}^2(2x)\,\mathrm dx&=2\coth(2x) \end{aligned}

The integrand is undefined when sinh⁡(2x)=0\sinh(2x)=0, namely at x=0x=0.

Answer. 2cosh⁡(3x)+2coth⁡(2x)+C2\cosh(3x)+2\coth(2x)+C on either interval x<0x<0 or x>0x>0. The constants on those separate intervals need not agree.

Check. Differentiate the coth term directly as 2cosh⁡(2x)/sinh⁡(2x)2\cosh(2x)/\sinh(2x). The quotient rule gives

4sinh⁡2(2x)−cosh⁡2(2x)sinh⁡2(2x)=−4cosech⁡2(2x),4\frac{\sinh^2(2x)-\cosh^2(2x)}{\sinh^2(2x)} =-4\operatorname{cosech}^2(2x),

confirming the sign independently of the memorised integral.

The same reverse-chain reasoning works for a non-linear argument whenever its derivative is present: ∫2xcosh⁡(x2) dx=sinh⁡(x2)+C\int2x\cosh(x^2)\,\mathrm dx=\sinh(x^2)+C. A missing inner derivative cannot be supplied by guesswork.

Watch out

∫sech⁡2(2x) dx=2tanh⁡(2x)+C\int\operatorname{sech}^2(2x)\,\mathrm dx=2\tanh(2x)+C is wrong: differentiating it gives four times the integrand. The correct coefficient is 1/21/2.

Use both the scale and sign decisions in a fresh integral.

Recognising inverse-function forms

A square root in the denominator can signal an inverse function. For example,

ddxsin⁡−1(x3)=1/31−x2/9=19−x2(−3<x<3).\frac{\mathrm d}{\mathrm dx}\sin^{-1}\left(\frac x3\right) =\frac{1/3}{\sqrt{1-x^2/9}} =\frac1{\sqrt{9-x^2}} \quad(-3<x<3).

The factor 1/31/3 cancels the factor extracted from the positive square root. There is no extra factor 33 in this antiderivative.

Three inverse square-root forms

For a>0a>0,

∫dxa2−x2=sin⁡−1(xa)+C(∣x∣<a)∫dxx2+a2=sinh⁡−1(xa)+C(x∈R)∫dxx2−a2=cosh⁡−1(xa)+C(x>a)\begin{aligned} \int\frac{\mathrm dx}{\sqrt{a^2-x^2}} &=\sin^{-1}\left(\frac xa\right)+C &&(|x|<a) \\ \int\frac{\mathrm dx}{\sqrt{x^2+a^2}} &=\sinh^{-1}\left(\frac xa\right)+C &&(x\in\mathbb R) \\ \int\frac{\mathrm dx}{\sqrt{x^2-a^2}} &=\cosh^{-1}\left(\frac xa\right)+C &&(x>a) \end{aligned}

All three forms, with these branch restrictions, are in the formula booklet.

The sign pattern decides the inverse: constant minus square gives inverse sine; square plus constant gives inverse sinh; square minus constant gives inverse cosh on its positive branch.

Worked example 3: a sum of squares

Question. Find ∫dxx2+9\displaystyle\int\frac{\mathrm dx}{\sqrt{x^2+9}}.

Plan. Match 9=329=3^2 to the inverse sinh form.

Here a=3>0a=3>0, and x2+9x^2+9 is positive for every real xx. Direct recognition gives the result.

Answer. sinh⁡−1(x/3)+C\sinh^{-1}(x/3)+C, for x∈Rx\in\mathbb R.

Check. The logarithmic form is ln⁡(x+x2+9)−ln⁡3\ln(x+\sqrt{x^2+9})-\ln3. Writing s=x2+9s=\sqrt{x^2+9}, its derivative is

1+x/sx+s=1s.\frac{1+x/s}{x+s}=\frac1s.

The logarithm is real even for negative xx, because s>∣x∣s>|x|.

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