Trigonometric and Hyperbolic Substitution

Everyone has a finite list in their head of functions they know how to integrate. Integration by substitution is a means of reducing an unfamiliar integral to a known integral.

For example, the integral

\displaystyle I=\int\frac{\ln x}{x}\,dx

wouldn’t look familiar to most people, but if we make the substitution x=e^y, this integral I can be shown to be equivalent to

\displaystyle I=\int y\,dy

and everyone likes integrating polynomials.

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In some cases it’s useful to substitute x with a trigonometric or a hyperbolic function. A substitution of this kind is most useful when the integrand contains an element of the form

\pm(b^2\pm x^2)

where b is some real number.

We’re going to look at the following list of integrals which are very easy to confuse with one another, but each member of which is of the appropriate form to be simplified by either trigonometric or hyperbolic substitution:

\displaystyle \int\frac{dx}{\sqrt{b^2-x^2}},\,\int\frac{dx}{\sqrt{x^2-b^2}},\,\int\frac{dx}{\sqrt{b^2+x^2}},\,\int\frac{dx}{b^2-x^2},\,\int\frac{dx}{b^2+x^2}

For all their superficial similarities, each integral must be considered individually.

The use of trigonometric and hyperbolic substitution requires a fair level of familiarity with properties of the relevant functions, particularly their relation to one another and their derivatives. The most important identities to remember are

\cos^2\theta+\sin^2\theta=1

\cosh^2\theta-\sinh^2\theta=1

Let’s get started.

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The first integral is

\displaystyle I_1=\int\frac{dx}{\sqrt{b^2-x^2}}

It is worth mentioning that the integrand is not defined for |x|>b, because the root in the denominator is undefined if b^2-x^2\le 0. Therefore we’re only looking for a solution in a very narrow strip of the real number line.

We want to exploit the fact that

1-\sin^2\theta=\cos^2\theta

so the appropriate substitution is

x=b\sin\theta

To replace the differential dx we differentiate both sides of the expression:

\displaystyle \frac{dx}{d\theta}=b\cos\theta

dx=b\cos\theta\,d\theta

Hence

\displaystyle I_1=\int\frac{b\cos\theta\,d\theta}{\sqrt{b^2-b^2\sin^2\theta}}

A factor of b^2 can be pulled out of the root, which will cancel with the b on the integrand’s numerator. Then we’re left with

\displaystyle I_1=\int\frac{\cos\theta\,d\theta}{\sqrt{1-\sin^2\theta}}

From the trigonometric identity above, the contents of the root is equal to cosine squared. On taking the square root, it will cancel with the numerator, leaving

\displaystyle I_1=\int d\theta

hence

I_1=\theta+c

where c is a constant. In terms of x, the indefinite integral is

\displaystyle I_1=\arcsin\Big(\frac{x}{b}\Big)+c\quad\text{for}\quad|x|<b

The graph below shows the integrand in brown and the integral I_1 in gold. That was fairly painless, but don’t worry; it’s going to get worse.

i1The second integral is

\displaystyle I_2=\int\frac{dx}{\sqrt{x^2-b^2}}

This integral is defined only for |x|>b, because the root in the denominator of the integrand is undefined if x^2-b^2\le 0. So we’re considering the entire real number line apart from a narrow band around the origin.

Though superficially similar to I_1, the appropriate substitution to this integral is not trigonometric. There is a way you can tell this won’t work immediately. Say we tried x=b\sin\theta. Then since

|\sin\theta|\le 1

b|\sin\theta|\le b

|x|\le b

But this is precisely the region in which x is not defined! You will never be able to find a value of \theta which gives you a value of x in the domain in which the integrand is defined. If we go ahead and try the substitution, we’ll end up with

\displaystyle I_2=\int\frac{\cos\theta\,d\theta}{\sqrt{-\cos^2\theta}}

and since the denominator is never real, the integral is completely broken; the trigonometric substitution gets us nowhere.

The appropriate substitution is actually

x=\pm b\cosh\theta

Why the plus or minus? Remember that hyperbolic cosine is a strictly positive function:

\cosh\theta\ge 1\,\forall\,\theta

So if we want a solution for x>0, the substitution must be of the form x=b\cosh\theta, and if we want a solution for x<0, the substitution must be of the form x=-b\cosh\theta.

Let’s take a solution for positive x:

x=b\cosh\theta

dx=b\sinh\theta\,d\theta

\displaystyle I_2=\int\frac{b\sinh\theta\,d\theta}{\sqrt{b^2\cosh^2\theta-b^2}}

Exploiting the fact that

\cosh^2\theta-1=\sinh^2\theta

we are left with

\displaystyle I_2=\int d\theta

\displaystyle I_2=\text{arccosh}\Big(\frac{x}{b}\Big)+c\quad\text{for}\quad x>b

If we carry out the substitution and integration for negative x, we get

\displaystyle I_2=\text{arccosh}\Big(-\frac{x}{b}\Big)+c\quad\text{for}\quad x<-b

This piecewise definition can be compacted by saying

\displaystyle I_2=\text{sgn}(x)\,\text{arccosh}\Big|\frac{x}{b}\Big|+c\quad\text{for}\quad |x|>b

where \text{sgn}(x) is the sign of x.

As you can see, even though I_1 and I_2 look quite similar, the latter must be treated more carefully than the former. This is because we have to pay attention to whether the substitution will actually work in the domain over which the integrand is defined.

This wasn’t a problem for I_1 because the integrand was defined for |x|<b, and our substitution for x was

x=b\sin\theta

This was perfectly fine, because x never had to be any bigger than 1, and \sin\theta can be assume both positive and negative values – there is never a value of x for which an appropriate value of \theta cannot be found.

You might also have wondered why I didn’t use the substitution x=b\cos\theta for I_1. The reason should now be clear. Since |x|<b, |\theta|<\frac{\pi}{2}. But because \cos\theta\ge 0 in this domain, we would have had to worry about substitutions for two different domains again:

x=b\cos\theta\quad\text{for}\quad x>0

x=-b\cos\theta\quad\text{for}\quad x<0

The graph below shows the integrand in brown and the integral I_2 in gold.

i2The third integral is

\displaystyle I_3=\int\frac{dx}{\sqrt{b^2+x^2}}

This integral is a bit safer because the integrand is defined for all real x. This is because the contents of the root is never less than or equal to 0. This means we don’t need to worry about solutions in different domains.

We want to exploit the fact that

1+\sinh^2\theta=\cosh^2\theta

so the appropriate substitution is

x=b\sinh\theta

Our substitution works perfectly, because for every value of x we can find a real value of \theta. The algebra:

dx=b\cosh\theta\,d\theta

\displaystyle I_3=\int\frac{b\cosh\theta\,d\theta}{\sqrt{b^2+b^2\sinh^2\theta}}

\displaystyle I_3=\int\frac{\cosh\theta}{\sqrt{\cosh^2\theta}}\,d\theta

\displaystyle I_3=\int d\theta

I_3=\theta+c

\displaystyle I_3=\text{arcsinh}\Big(\frac{x}{b}\Big)+c

The graph below shows the integrand in brown and the integral I_3 in gold.

i3The fourth integral is

\displaystyle I_4=\int\frac{dx}{b^2-x^2}

We’ve now got no square root to worry about, so there are only two isolated places where the integrand is undefined (where x=\pm b). But we do have to worry about two different domains, specifically where |x|>b and where |x|<b.

Ideally, we want to exploit the facts that

1-\tanh^2\theta=\text{sech}^2\theta

1-\coth^2\theta=-\text{csch}^2\theta

So do we want to use \tanh\theta, or \coth\theta? The answer will become clear depending on which domain we’re working in.

First take the case where |x|<b. Then the appropriate substitution is

x=b\tanh\theta

since b|\coth\theta|\ge b, and would therefore never work as a substitution in this domain. We differentiate this expression:

\displaystyle \frac{dx}{d\theta}=b\frac{d}{d\theta}\Big(\frac{\sinh\theta}{\cosh\theta}\Big)

\displaystyle \frac{dx}{d\theta}=b\Big(\frac{\cosh^2\theta-\sinh^2\theta}{\cosh^2\theta}\Big)

\displaystyle \frac{dx}{d\theta}=b\,\text{sech}^2\theta

dx=b\,\text{sech}^2\theta\,d\theta

Using this substitution yields

\displaystyle I_4=\int\frac{b\,\text{sech}^2\theta\,d\theta}{b^2-b^2\tanh^2\theta}

Using the identity above, this collapses to

\displaystyle I_4=\frac{1}{b}\int d\theta

\displaystyle I_4=\frac{1}{b}\theta+c

\displaystyle I_4=\frac{1}{b}\text{arctanh}\Big(\frac{x}{b}\Big)+c\quad\text{for}\quad |x|<b

In the domain where |x|>b, the appropriate substitution is

x=b\coth\theta

since b|\tanh\theta|\le b, and would therefore never work in this domain. The algebra in this case gives

\displaystyle I_4=\frac{1}{b}\text{arccoth}\Big(\frac{x}{b}\Big)+c\quad\text{for}\quad |x|>b

The graph below shows the integrand in brown and the integral I_4 in gold.

i4The fifth integral is

\displaystyle I_5=\int\frac{dx}{b^2+x^2}

This is another relatively friendly integral, because its integrand is defined everywhere on the real number line, and its solution can be expressed as a single continuous function.

We want to exploit the fact that

1+\tan^2\theta=\sec^2\theta

so the appropriate substitution is

x=b\tan\theta

Just like \sinh\theta, \tan\theta can assume any value on the real number line, be it positive or negative, so we can use this substitution without worrying about the sign or size of x.

dx=b\sec^2\theta\,d\theta

\displaystyle I_5=\int\frac{b\sec^2\theta\,d\theta}{b^2\tan^2\theta+b^2}

\displaystyle I_5=\frac{1}{b}\int d\theta

\displaystyle I_5=\frac{1}{b}\arctan\Big(\frac{x}{b}\Big)+c

The graph below shows the integrand in brown and the integral I_5 in gold.

i5\

Here’s the full list of solutions:

\displaystyle \int\frac{dx}{\sqrt{b^2-x^2}}=\begin{cases}\arcsin\big(\frac{x}{b}\big)+c\quad\text{for}\quad|x|<b\\\text{undefined}\quad\text{for}\quad |x|\ge b\end{cases}

\displaystyle \int\frac{dx}{\sqrt{x^2-b^2}}=\begin{cases}\text{arccosh}\big(\frac{x}{b}\big)+c\quad\text{for}\quad x>b\\ -\text{arccosh}\big(-\frac{x}{b}\big)+c\quad\text{for}\quad x<-b\\ \text{undefined}\quad\text{for}\quad |x|\le b\end{cases}

\displaystyle \int\frac{dx}{\sqrt{b^2+x^2}}=\text{arcsinh}\Big(\frac{x}{b}\Big)+c\quad\forall\quad x

\displaystyle \int\frac{dx}{b^2-x^2}=\begin{cases}\frac{1}{b}\text{arctanh}\big(\frac{x}{b}\big)+c\quad\text{for}\quad |x|<b\\ \frac{1}{b}\text{arccoth}\big(\frac{x}{b}\big)+c\quad\text{for}\quad |x|>b\\ \text{undefined}\quad\text{for}\quad x=\pm b\end{cases}

\displaystyle \int\frac{dx}{b^2+x^2}=\frac{1}{b}\arctan\Big(\frac{x}{b}\Big)+c\quad\forall\quad x

The solutions are all somewhat alike, in that they are all arc-somethings of the variable \frac{x}{b}. They are not individually that difficult to remember, but telling which solution goes with which integral is tricky. The situation is particularly complex when it comes to the piecewise definition of certain solutions. I think this list shows it’s not worth trying to remember all the various solutions, and that becoming familiar with trigonometric and hyperbolic substitution is worthwhile.

You’ll notice that the integrands with square roots in them were solved using s(h)iney and cos(h)iney type functions, and those without roots were solved using t(h)any cot(h)y type functions.

You’ll also notice that we had to be much more careful with integrands containing \pm(x^2-b^2) than those containing x^2+b^2. You can expect that, if the integrand contains \pm(x^2-b^2), you will probably have to define your solution piecewise.

Perhaps the most important part of choosing a substitution is making sure your new variable is actually able to assume some real values! For example, if your integrand is defined for x>1 only, you probably shouldn’t choose the substitution x=\sin\theta.

The real question is: how did I know which substitution to do when? The answer: I didn’t. The best I could do was narrow the substitution down to a trigonometric or a hyperbolic one. To a great extent, integration by substitution is guesswork. You try a substitution; maybe it works, maybe it doesn’t. Sometimes the integrand gets simpler, sometimes you make an even bigger mess of it. The only way of improving is through experience. Eventually you’ll start to recognise certain forms. You might not be able to memorise the exact answer to an integral, but you may be able to recognise the appropriate substitution. That is: you don’t need to remember the solution, only how to find it.

To reiterate, trigonometric and hyperbolic substitutions are a fairly safe bet when you’ve got an integral containing

\pm(b^2\pm x^2)

Their scope goes beyond this, of course. For example, the substitution

x=b\sec^2\theta

is useful for solving an integral that will be forever burned into my memory:

\displaystyle \int\frac{dx}{\sqrt{1-\frac{b}{x}}}

There are even funny half-angle substitutions for solving integrals like

\displaystyle \int\frac{d\varphi}{a+b\cos\varphi}

As with everything in maths, it’s just a matter of practice. Or divine inspiration.

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