2.4 - Trigonometric and hyperbolic substitutions | Cambridge A-Level Further Mathematics

A substitution can replace a square root by a trigonometric or hyperbolic identity. You will choose substitutions from the sign of a quadratic, keep the correct real branch and transform limits to obtain exact definite integrals.

Sine and tangent substitutions

How can a denominator such as $(9+x^2)^{3/2}$ become easier rather than harder? Choose $x=3\tan\theta$ : the identity $1+\tan^2\theta=\sec^2\theta$ turns the square root into $3\sec\theta$ , while $\mathrm dx=3\sec^2\theta\,\mathrm d\theta$ cancels two of its three secant factors.

A substitution is useful when it simplifies the whole expression, including the differential. For a constant minus a square, use sine instead: $x=a\sin\theta$ gives $a^2-x^2=a^2\cos^2\theta$ .

Two radical substitutions

For $a>0$ , choose branches with positive cosine:

\begin{array}{c|c|c} \text{Radical}&\text{Substitution}&\text{Differential and root}\\ \hline \sqrt{a^2-x^2}&x=a\sin\theta&\mathrm dx=a\cos\theta\,\mathrm d\theta,\quad\sqrt{a^2-x^2}=a\cos\theta\\ \sqrt{a^2+x^2}&x=a\tan\theta&\mathrm dx=a\sec^2\theta\,\mathrm d\theta,\quad\sqrt{a^2+x^2}=a\sec\theta \end{array}

Use $-\pi/2<\theta<\pi/2$ for these identities without absolute values. These substitution rules are not listed as such in the formula booklet: derive them from the trigonometric identities and chain rule.

The chosen branch matters because $\sqrt{\cos^2\theta}=|\cos\theta|$ . We may remove the absolute value only after ensuring $\cos\theta>0$ .

Worked example 1: sine removes a difference of squares

Question. Find $\displaystyle\int\frac{\mathrm dx}{(9-x^2)^{3/2}}$ , for $-3<x<3$ .

Plan. Set $x=3\sin\theta$ , so the positive square root becomes $3\cos\theta$ .

On the principal sine branch, $\mathrm dx=3\cos\theta\,\mathrm d\theta$ . Hence

\begin{aligned} \int\frac{\mathrm dx}{(9-x^2)^{3/2}} &=\frac19\int\sec^2\theta\,\mathrm d\theta\\ &=\frac19\tan\theta+C \end{aligned}

Since $\sin\theta=x/3$ and $\cos\theta=\sqrt{9-x^2}/3$ , their ratio gives the answer in $x$ .

Answer. $\displaystyle\frac{x}{9\sqrt{9-x^2}}+C$ , for $|x|<3$ .

Check. Differentiate $x(9-x^2)^{-1/2}/9$ . The numerator after taking a common denominator is $(9-x^2)+x^2=9$ , giving exactly $(9-x^2)^{-3/2}$ .

Change only the minus sign inside the denominator to a plus. Tangent now fits the identity; using sine would not remove the sum of squares.

Worked example 2: tangent removes a sum of squares

Question. Find $\displaystyle\int\frac{\mathrm dx}{(9+x^2)^{3/2}}$ .

Plan. Set $x=3\tan\theta$ , choosing $-\pi/2<\theta<\pi/2$ .

Then $\mathrm dx=3\sec^2\theta\,\mathrm d\theta$ and $(9+x^2)^{3/2}=27\sec^3\theta$ , so

\begin{aligned} \int\frac{\mathrm dx}{(9+x^2)^{3/2}} &=\frac19\int\cos\theta\,\mathrm d\theta\\ &=\frac19\sin\theta+C \end{aligned}

Here $\tan\theta=x/3$ and $\sec\theta=\sqrt{9+x^2}/3$ . Thus $\sin\theta=\tan\theta/\sec\theta=x/\sqrt{9+x^2}$ , including negative $x$ .

Answer. $\displaystyle\frac{x}{9\sqrt{9+x^2}}+C$ , for every real $x$ .

Check. Direct differentiation gives

\frac19\frac{(9+x^2)-x^2}{(9+x^2)^{3/2}} =\frac1{(9+x^2)^{3/2}}.

The answer's derivative is positive and even, just like the original integrand.

A definite integral may be evaluated entirely in the new variable. For an indefinite integral, convert back to the original variable, as both examples did.

Watch out

Replacing $x$ inside the radical while leaving $\mathrm dx$ unchanged is not a substitution. Write the differential first; it often supplies the factor that makes the method work.

Use the tangent substitution with a different scale.

Sinh and cosh substitutions

The identity $\cosh^2u-\sinh^2u=1$ offers another way to remove a sum of squares. Setting $x=2\sinh u$ makes $\sqrt{x^2+4}=2\cosh u$ , with no absolute-value issue because cosh is always positive.

For a difference of squares on $x>2$ , set $x=2\cosh u$ and choose $u>0$ . Then $\sqrt{x^2-4}=2\sinh u$ . The restriction makes sinh positive and makes the substitution one-to-one.

Hyperbolic radical substitutions

For $a>0$ ,

\begin{array}{c|c|c} \text{Radical}&\text{Substitution}&\text{Differential and root}\\ \hline \sqrt{x^2+a^2}&x=a\sinh u&\mathrm dx=a\cosh u\,\mathrm du,\quad\sqrt{x^2+a^2}=a\cosh u\\ \sqrt{x^2-a^2},\ x>a&x=a\cosh u,\ u>0&\mathrm dx=a\sinh u\,\mathrm du,\quad\sqrt{x^2-a^2}=a\sinh u \end{array}

These substitution rules are not printed as such in the formula booklet. The fundamental and double-angle hyperbolic identities used to derive them are supplied there.

A radical in the numerator becomes a square of a hyperbolic function after including the differential. Use $\cosh^2u=(\cosh2u+1)/2$ or $\sinh^2u=(\cosh2u-1)/2$ , derived from the supplied identities.

Worked example 3: integrating a square root of a sum

Question. Use a hyperbolic substitution to find $\displaystyle\int\sqrt{x^2+4}\,\mathrm dx$ .

Plan. Set $x=2\sinh u$ , then use a double-angle identity to integrate $\cosh^2u$ .

For any real $u$ ,

\begin{aligned} \int\sqrt{x^2+4}\,\mathrm dx &=4\int\cosh^2u\,\mathrm du\\ &=2\int(\cosh2u+1)\,\mathrm du\\ &=\sinh2u+2u+C \end{aligned}

Now $u=\sinh^{-1}(x/2)$ , and

\sinh2u=2\sinh u\cosh u =\frac{x\sqrt{x^2+4}}2.

Answer. $\tfrac12x\sqrt{x^2+4}+2\sinh^{-1}(x/2)+C$ , for every real $x$ .

Check. With $s=\sqrt{x^2+4}$ , differentiation gives

\frac s2+\frac{x^2}{2s}+\frac2s =\frac{s^2+x^2+4}{2s}=s,

which verifies the entire antiderivative without repeating the substitution.

Change only the plus sign to a minus and restrict to $x>2$ . Cosh replaces sinh in the substitution, while the double-angle identity changes the sign of the inverse term.

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2.4 - Trigonometric and hyperbolic substitutions

Sine and tangent substitutions

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