26.7 Further Integration by Substitution
AHL 5.16
This section is for HL students. It builds on 26.6.
- how to use a given substitution when the derivative does not cancel, by writing \(x\) in terms of \(u\)
- how to use a given substitution for harder integrals, including \(u = \sqrt{\ldots}\) and \(x = a\sin\theta\)
At HL the substitution is usually given in the question. Use the four steps of 26.6.
Writing \(x\) in terms of \(u\)
Sometimes the derivative does not cancel. Then write \(x\) itself in terms of \(u\).
Example 1 By using the substitution \(u = x - 3\), find \(\displaystyle\int x(x - 3)^4\,\mathrm{d}x\).
\(\mathrm{d}u = \mathrm{d}x\) and \(x = u + 3\).
\[ \int (u + 3)u^4\,\mathrm{d}u = \int \left(u^5 + 3u^4\right)\mathrm{d}u = \frac{u^6}{6} + \frac{3u^5}{5} + C \]
\[ = \frac{(x - 3)^6}{6} + \frac{3(x - 3)^5}{5} + C. \]
(Expanding \(x(x - 3)^4\) first also works, but takes much longer.)
Harder substitutions
Example 2 By using the substitution \(u = \sqrt{x + 4}\), show that
\[ \int \frac{1}{x\sqrt{x + 4}}\,\mathrm{d}x = \frac{1}{2}\ln\left\lvert\frac{\sqrt{x + 4} - 2}{\sqrt{x + 4} + 2}\right\rvert + C. \]
\(u = \sqrt{x + 4}\), so \(u^2 = x + 4\) and \(x = u^2 - 4\). Then \(\mathrm{d}x = 2u\,\mathrm{d}u\).
\[ \int \frac{1}{\left(u^2 - 4\right)u}\,2u\,\mathrm{d}u = \int \frac{2}{u^2 - 4}\,\mathrm{d}u = \int \frac{2}{(u - 2)(u + 2)}\,\mathrm{d}u. \]
Partial fractions: \(2 = A(u + 2) + B(u - 2)\). \(\;u = 2\): \(A = \dfrac{1}{2}\). \(\;u = -2\): \(B = -\dfrac{1}{2}\).
\[ \int \left(\frac{\frac{1}{2}}{u - 2} - \frac{\frac{1}{2}}{u + 2}\right)\mathrm{d}u = \frac{1}{2}\ln\left\lvert\frac{u - 2}{u + 2}\right\rvert + C = \frac{1}{2}\ln\left\lvert\frac{\sqrt{x + 4} - 2}{\sqrt{x + 4} + 2}\right\rvert + C. \quad\blacksquare \]
When \(\sqrt{a^2 - x^2}\) appears, the substitution \(x = a\sin\theta\) removes the square root, because \(a^2 - a^2\sin^2\theta = a^2\cos^2\theta\).
Example 3 By using the substitution \(x = 3\sin\theta\), where \(-\dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}\), find \(\displaystyle\int \sqrt{9 - x^2}\,\mathrm{d}x\).
\(\mathrm{d}x = 3\cos\theta\,\mathrm{d}\theta\). Also \(\sqrt{9 - 9\sin^2\theta} = \sqrt{9\cos^2\theta} = 3\cos\theta\) (since \(\cos\theta \geq 0\) for this range of \(\theta\)).
\[ \int 3\cos\theta \cdot 3\cos\theta\,\mathrm{d}\theta = 9\int \cos^2\theta\,\mathrm{d}\theta = \frac{9}{2}\int (1 + \cos 2\theta)\,\mathrm{d}\theta = \frac{9}{2}\theta + \frac{9}{4}\sin 2\theta + C. \]
Return to \(x\). \(\sin\theta = \dfrac{x}{3}\), so \(\theta = \arcsin\dfrac{x}{3}\) and \(\cos\theta = \dfrac{\sqrt{9 - x^2}}{3}\). Then
\[ \sin 2\theta = 2\sin\theta\cos\theta = \frac{2x\sqrt{9 - x^2}}{9}. \]
\[ \int \sqrt{9 - x^2}\,\mathrm{d}x = \frac{9}{2}\arcsin\left(\frac{x}{3}\right) + \frac{x}{2}\sqrt{9 - x^2} + C. \]
A quick check of Example 3: the definite integral from \(0\) to \(3\) must equal \(F(3) - F(0) = \dfrac{9}{2} \cdot \dfrac{\pi}{2} = \dfrac{9\pi}{4} \approx 7.07\).
Calculator → Menu → Calculus → Numerical Integral, then enter \(\displaystyle\int_0^3 \sqrt{9 - x^2}\,\mathrm{d}x\). The GDC gives \(7.0685\ldots\), which matches. (It is also a quarter of a circle of radius 3.) Definite integrals are studied in Unit 27.
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| Integrating while \(x\) and \(u\) are mixed | Write every \(x\) in terms of \(u\) first | You cannot integrate \(x\) with respect to \(u\). |
| \(\sqrt{9\cos^2\theta} = \pm 3\cos\theta\) | \(3\cos\theta\) | The given range makes \(\cos\theta \geq 0\). |
Exercises
1 By using the substitution \(u = 2x + 1\), find \(\displaystyle\int \frac{x}{\sqrt{2x + 1}}\,\mathrm{d}x\).
\(x = \dfrac{u - 1}{2}\), \(\mathrm{d}x = \dfrac{1}{2}\mathrm{d}u\):
\(\displaystyle\int \frac{u - 1}{2\sqrt{u}} \cdot \frac{1}{2}\,\mathrm{d}u = \frac{1}{4}\int \left(u^{\frac{1}{2}} - u^{-\frac{1}{2}}\right)\mathrm{d}u = \frac{1}{6}u^{\frac{3}{2}} - \frac{1}{2}u^{\frac{1}{2}} + C\).
Taking out \(\dfrac{1}{6}\sqrt{u}\): \(\;\dfrac{1}{6}\sqrt{u}\,(u - 3) = \dfrac{1}{6}\sqrt{2x + 1}\,(2x - 2) = \dfrac{1}{3}(x - 1)\sqrt{2x + 1} + C\).
2 By using the substitution \(x = 2\tan\theta\), where \(-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}\), find \(\displaystyle\int \frac{1}{\left(4 + x^2\right)^{\frac{3}{2}}}\,\mathrm{d}x\).
\(\mathrm{d}x = 2\sec^2\theta\,\mathrm{d}\theta\) and \(4 + 4\tan^2\theta = 4\sec^2\theta\), so \(\left(4 + x^2\right)^{\frac{3}{2}} = 8\sec^3\theta\).
\(\displaystyle\int \frac{2\sec^2\theta}{8\sec^3\theta}\,\mathrm{d}\theta = \frac{1}{4}\int \cos\theta\,\mathrm{d}\theta = \frac{1}{4}\sin\theta + C\).
From \(\tan\theta = \dfrac{x}{2}\), a right-angled triangle gives \(\sin\theta = \dfrac{x}{\sqrt{4 + x^2}}\). Answer: \(\dfrac{x}{4\sqrt{4 + x^2}} + C\).
3 By using the substitution \(u = \sin x\), find \(\displaystyle\int \frac{\cos x}{\sin^2 x + 3\sin x + 2}\,\mathrm{d}x\).
\(\mathrm{d}u = \cos x\,\mathrm{d}x\): \(\;\displaystyle\int \frac{1}{(u + 1)(u + 2)}\,\mathrm{d}u = \int \left(\frac{1}{u + 1} - \frac{1}{u + 2}\right)\mathrm{d}u = \ln\left\lvert\frac{u + 1}{u + 2}\right\rvert + C\).
\(= \ln\left(\dfrac{\sin x + 1}{\sin x + 2}\right) + C\) (the integrand needs \(\sin x \neq -1\), so \(\sin x + 1 > 0\); and \(\sin x + 2 > 0\) always).