This note shows how to integrate composite expressions by substitution, which reverses the chain rule, and how to find the volume of a solid formed by rotating a curve about the \(x\)-axis. These methods turn awkward integrals into simple ones and let you model the volume of objects such as bottles, bowls and lenses.
Key ideas
Why substitution works. The chain rule differentiates \(F(g(x))\) to give \(F'(g(x))\,g'(x)\). Integration by substitution runs this backwards. If an integrand contains a function \(g(x)\) together with (a multiple of) its derivative \(g'(x)\), you can replace \(g(x)\) with a new variable \(u\) and make the integral much simpler.
The method.
- Choose \(u\) as the "inside" part, often the expression in brackets, under a root, or in an exponent. In many questions the substitution is given to you.
- Differentiate: \(\dfrac{du}{dx}=u'(x)\), then write \(du=u'(x)\,dx\).
- Rewrite the whole integral in terms of \(u\) and \(du\). No \(x\) may remain.
- Integrate with respect to \(u\), then replace \(u\) by its expression in \(x\) for an indefinite integral.
Definite integrals. You have two choices. You can find the antiderivative, return to \(x\), and use the original limits. Or you can change the limits: substitute each \(x\)-limit into \(u\) to get \(u\)-limits, then evaluate entirely in \(u\). The second method avoids converting back. Whichever you use, do not mix the two, because the \(x\)-limits must never be applied to an expression in \(u\).
Results worth knowing. If the numerator is the derivative of the denominator, then \(\int\dfrac{f'(x)}{f(x)}dx=\ln|f(x)|+c\). Also \(\int e^{g(x)}g'(x)\,dx=e^{g(x)}+c\). Both come straight from the chain rule.
Volumes of revolution. Rotate the region between a curve \(y=f(x)\), the \(x\)-axis and the lines \(x=a\) and \(x=b\) through a full turn (360°) about the \(x\)-axis. The solid is made of thin discs of radius \(y\) and thickness \(dx\). Each disc has volume \(\pi y^2\,dx\), and adding them all gives \[V=\pi\int_a^b y^2\,dx.\] Square the whole function before integrating, not each term separately. You can use the GDC to evaluate the integral when it has no neat antiderivative.
Choosing \(u\). Ask yourself which part of the integrand, when differentiated, gives something already present as a factor. In \(x(x^2+4)^5\), the bracket \(x^2+4\) differentiates to \(2x\), and there is an \(x\) outside, so the bracket is the right choice. A constant factor missing is not a problem, because you can balance it, as in Example 1. A missing power of \(x\) is a problem, and means a different approach is needed.
Checking. Differentiate your answer using the chain rule. If you get back the original integrand, the integration is correct. For a definite integral, a GDC value is a quick check.
Describing the solid. The volume is a number of cubic units, so state the units in context. If a region lies partly below the \(x\)-axis, squaring \(y\) makes every disc contribute a positive volume, so no splitting is needed for the volume formula.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(du=u'(x)\,dx\) | Links the new variable to \(dx\) in a substitution. |
| \(\int f(g(x))\,g'(x)\,dx=\int f(u)\,du\) | Reverse chain rule with \(u=g(x)\). |
| \(\int\dfrac{f'(x)}{f(x)}dx=\ln|f(x)|+c\) | Top is the derivative of the bottom. |
| \(\int e^{g(x)}g'(x)\,dx=e^{g(x)}+c\) | Exponential with the inner derivative present. |
| \(V=\pi\int_a^b y^2\,dx\) | Volume when the region under \(y=f(x)\) rotates about the \(x\)-axis. |
Worked examples
Example 1 Find \(\int x(x^2+4)^5\,dx\) using the substitution \(u=x^2+4\).
\(\dfrac{du}{dx}=2x\), so \(du=2x\,dx\) and \(x\,dx=\dfrac12du\).
The integral becomes \(\int u^5\cdot\dfrac12\,du=\dfrac12\cdot\dfrac{u^6}{6}+c=\dfrac{u^6}{12}+c\).
Replace \(u\): \(\dfrac{(x^2+4)^6}{12}+c\).
Answer: \(\dfrac{(x^2+4)^6}{12}+c\)
Example 2 Evaluate \(\int_0^{\pi/2}\sin x\cos^2x\,dx\) using \(u=\cos x\).
\(\dfrac{du}{dx}=-\sin x\), so \(\sin x\,dx=-du\).
Change the limits: when \(x=0\), \(u=\cos 0=1\). When \(x=\dfrac\pi2\), \(u=\cos\dfrac\pi2=0\).
The integral becomes \(\int_1^0 u^2(-du)=\int_0^1u^2\,du\) (swapping the limits removes the minus sign).
\(=\left[\dfrac{u^3}{3}\right]_0^1=\dfrac13\).
Answer: \(\dfrac13\)
Example 3 Find the exact value of \(\int_0^1\dfrac{2x}{x^2+1}\,dx\).
The numerator \(2x\) is the derivative of the denominator \(x^2+1\), so the integral is \(\ln(x^2+1)\).
(Using \(u=x^2+1\), \(du=2x\,dx\) gives \(\int\dfrac{du}{u}=\ln|u|\), the same result.)
\([\ln(x^2+1)]_0^1=\ln2-\ln1=\ln 2\approx 0.693\).
Answer: \(\ln 2\approx 0.693\)
Example 4 A decorative vase is modelled by rotating the curve \(y=x^2+1\), for \(0\le x\le 1\) (units in decimetres), through 360° about the \(x\)-axis. Find its volume.
\(V=\pi\int_0^1(x^2+1)^2dx\).
Expand first: \((x^2+1)^2=x^4+2x^2+1\).
Integrate: \(\left[\dfrac{x^5}{5}+\dfrac{2x^3}{3}+x\right]_0^1=\dfrac15+\dfrac23+1=\dfrac{3+10+15}{15}=\dfrac{28}{15}\).
So \(V=\dfrac{28\pi}{15}\approx 5.86\).
Answer: \(\dfrac{28\pi}{15}\approx 5.86\) cubic decimetres
Example 5 The region under \(y=e^{-x^2}\), for \(0\le x\le 1\), is rotated about the \(x\)-axis. Find the volume.
\(V=\pi\int_0^1\left(e^{-x^2}\right)^2dx=\pi\int_0^1e^{-2x^2}dx\).
This has no simple antiderivative, so use the definite integral feature of your GDC: \(\int_0^1e^{-2x^2}dx\approx 0.5981\).
Multiply by \(\pi\): \(V\approx 1.879\).
Answer: \(V\approx 1.88\) cubic units
Common errors
- Leaving some \(x\) in the integral after substituting. If the \(x\) cannot be removed, the substitution is not suitable.
- Forgetting the constant factor from \(du\), such as writing \(\int u^5du\) instead of \(\dfrac12\int u^5du\) in Example 1.
- Using the \(x\)-limits on an expression in \(u\). Either change the limits or return to \(x\) before evaluating.
- Squaring each term of \(y\) separately in a volume, for example writing \((x^2+1)^2\) as \(x^4+1\).
- Forgetting the factor \(\pi\), or leaving the answer with no units when the question gives units.
Exam tip: After a substitution, check that every part of the integrand has been converted, including the limits. When you quote a volume, give both the exact form with \(\pi\) (when it is easy to find) and a decimal to 3 s.f.
Check yourself
- Find \(\int 3x^2(x^3+2)^4\,dx\).
Answer
Let \(u=x^3+2\), so \(du=3x^2dx\). Then \(\int u^4du=\dfrac{u^5}{5}+c=\dfrac{(x^3+2)^5}{5}+c\). - Evaluate \(\int_0^1xe^{x^2}\,dx\).
Answer
Let \(u=x^2\), so \(x\,dx=\dfrac12du\); limits go from 0 to 1. \(\dfrac12\int_0^1e^udu=\dfrac12(e-1)\approx 0.859\). - Evaluate \(\int_0^{\pi/2}\cos x\sin^3x\,dx\).
Answer
Let \(u=\sin x\), so \(du=\cos x\,dx\); limits go from 0 to 1. \(\int_0^1u^3du=\dfrac14\). - The region under \(y=\sqrt{9-x^2}\), for \(-3\le x\le 3\), is rotated about the \(x\)-axis. Find the volume.
Answer
\(V=\pi\int_{-3}^{3}(9-x^2)dx=\pi\left[9x-\dfrac{x^3}{3}\right]_{-3}^{3}=\pi\left[(27-9)-(-27+9)\right]=36\pi\approx 113\). This is a sphere of radius 3.