The term “integration” is used in two ways. It refers to:

• The process of finding areas or differences of areas (computing a definite integral), and also
• The process of finding an antiderivative (evaluating an indefinite integral).

REMINDER A “composite function” is a function of the form \(f(g(x))\). For convenience, we call \(g(x)\) the inside function and \(f(u)\) the outside function.

THEOREM 1 The Substitution Method

If \(F'(x) = f(x)\), then \[ \boxed{\bbox[#FAF8ED,5pt]{\displaystyle \int f(u(x)) u'(x)\, \mathrm{d}x = F(u(x)) + C}} \]

The symbolic calculus of substitution using differentials was invented by Leibniz and is considered one of his most important achievements. It reduces the otherwise complicated process of transforming integrals to a convenient set of rules.

EXAMPLE 1

Evaluate \(\displaystyle \int 3x^2\sin(x^3)\, \mathrm{d}x\).

Solution The integrand contains the composite function \(sin(x^{3})\), so we set \(u = x^{3}\). The differential \(\mathrm{d}u = 3x^{2}\, \mathrm{d}x\) also appears, so we can carry out the substitution: \[ \int 3x^2\sin(x^3)\, \mathrm{d}x = \int \underbrace{\sin(x^3)}_{\sin(u)}\underbrace{3x^2\, \mathrm{d}x}_{\mathrm{d}u} = \int \sin u\, \mathrm{d}u \]

Now evaluate the integral in the \(u\)-variable and replace \(u\) by \(x^{3}\) in the answer: \[ \int 3x^2\sin(x^3)\, \mathrm{d}x =\int \sin u\, \mathrm{d}u = -\cos u +C = -\cos(x^3) + C \]

Let’s check our answer by differentiating: \[ \dfrac{d}{dx} (-\cos(x^3)) = \sin(x^3)\dfrac{d}{dx}x^3 = 3x^2 \sin(x^3) \]

EXAMPLE 2 Multiplying \(\mathrm{d}u\) by a Constant

Evaluate \(\displaystyle \int x(x^2+9)^5 \mathrm{d}x\).

Solution We let \(u = x^{2} + 9\) because the composite \(u^{5} = (x^{2} + 9)^{5}\) appears in the integrand. The differential \(\mathrm{d}u = 2x\, \mathrm{d}x\) does not appear as is, but we can multiply by \(\tfrac12\) to obtain \[ \dfrac12\, \mathrm{d}u = x\, \mathrm{d}x \implies \dfrac12u^5\, \mathrm{d}u = x(x^2 + 9)^5\, \mathrm{d}x \] Now we can apply substitution: \[ \int x(x^2 + 9)^5\, \mathrm{d}x = \int \overbrace{(x^2+9)^5}^{u^5} \overbrace{x\, \mathrm{d}x}^{\tfrac12\, \mathrm{d}u} = \dfrac12 \int u^5\, \mathrm{d}u = \dfrac1{12}u^6 + C \] Finally, we express the answer in terms of \(x\) by substituting \(u = x^{2} + 9\): \[ \int x(x^2 + 9)^5\, \mathrm{d}x = \dfrac1{12}u^6 + C = \dfrac1{12}(x^2+9)^6 + C \]

Substitution Method:

1. Choose \(u\) and compute \(\mathrm{d}u\).
2. Rewrite the integral in terms of \(u\) and \(\mathrm{d}u\), and evaluate.
3. Express the final answer in terms of \(x\).

EXAMPLE 3

Evaluate \(\displaystyle \int \dfrac{(x^2 + 2x)\, \mathrm{d}x}{(x^3 + 3x^2 + 12)^6}\).

Solution The appearance of \((x^{3} + 3x^{2} + 12)^{-6}\) in the integrand suggests that we try \(u = x^{3} + 3x^{2} + 12\). With this choice, \[ \mathrm{d}u = (3x^2 + 6x)\, \mathrm{d}x = 3 (x^2 + 2x)\, \mathrm{d}x \implies \dfrac13 \mathrm{d}u = (x^2+2x)\, \mathrm{d}x \] \[ \begin{aligned} \int \dfrac{(x^2 + 2x)\, \mathrm{d}x}{(x^3 + 3x^2 + 12)^6} & = \int \overbrace{(x^3 + 3x^2 + 12)^{-6}}^{u^{-6}} \overbrace{(x^2 + 2x)\, \mathrm{d}x}^{\tfrac13\, \mathrm{d}u}\\ & =\dfrac13 \int u^{-6}\, \mathrm{d}u = \left(\dfrac13\right)\left(\dfrac{u^{-5}}{-5}\right) + C \\ & = - \dfrac1{15}(x^3 + 3x^2 + 12)^{-5} + C \end{aligned} \]

CONCEPTUAL INSIGHT

An integration method that works for a given function may fail if we change the function even slightly. In the previous example, if we replace 2 by 2.1 and consider instead \(\displaystyle \int\dfrac{(x^2 + 2.1x)\, \mathrm{d}x}{(x^3 + 3x^2 + 12)^6}\), the Substitution Method does not work. The problem is that \((x^{2} + 2.1x)\, \mathrm{d}x\) is not a multiple of \(\mathrm{d}u = (3x^{2} + 6x)\, \mathrm{d}x\).

EXAMPLE 4

Evaluate \(\displaystyle\int \sin(7\theta+5)\, \mathrm{d}\theta\).

Solution Let \(u = 7 \theta + 5\). Then \(\mathrm{d}u = 7\, \mathrm{d}\theta\) and \(\tfrac17\, \mathrm{d}u=\mathrm{d}\theta\). We obtain \[ \sin(7\theta+5)\, \mathrm{d}\theta = \dfrac17 \int \sin u\, \mathrm{d}u = -\dfrac17 \cos u + C = -\dfrac17\cos(7\theta + 5) + C \]

EXAMPLE 5

Evaluate \(\displaystyle \int e^{-9t}\, \mathrm{d}t\).

Solution Use the substitution \(u = -9t\), \(\mathrm{d}u = -9\, \mathrm{d}t\): \[ \int e^{-9t}\, \mathrm{d}t = \int e^{u}\left(-\dfrac19\, \mathrm{d}u\right) = -\dfrac19\int e^u\, \mathrm{d}u = -\dfrac19 e^u + C = -\dfrac19 e^{-9t} + C \]

EXAMPLE 6

Evaluate \(\displaystyle \int \tan\theta\, \mathrm{d}\theta\).

Solution In this case, the idea is to write \(\displaystyle \tan\theta\, \mathrm{d}\theta = \dfrac{\sin\theta\, \mathrm{d}\theta}{\cos\theta}\) and to note that if \(u = \cos\theta\). Then \(\mathrm{d}u = -\sin\theta\, \mathrm{d}\theta\) and \[ \int \tan\theta\, \mathrm{d}\theta = \int \dfrac{\sin\theta\, \mathrm{d}\theta}{\cos\theta} = - \int \dfrac{\mathrm{d}u}{u} = -\ln|u| + C = - \ln|\cos(\theta)| + C \] Now recall that \(-\ln u = \ln\tfrac 1u\). Thus, \(-\ln|\cos(\theta)| = \ln\tfrac1{|\cos(\theta)|}\) and we obtain \[ \int \tan\theta\, \mathrm{d}\theta =\ln\lvert\dfrac{1}{\cos(\theta)}\rvert + C = \ln|\sec\theta| + C \]

The substitution method does not always work, even when the integral looks relatively simple. For example, \(\int \sin(x^2)\, \mathrm{d}x\) cannot be evaluated explicitly by substitution, or any other method. With experience, you will learn to recognize when substitution is likely to be successful.

EXAMPLE 7 Additional Step Necessary

Evaluate \(\displaystyle \int x\sqrt{5x+1}\, \mathrm{d}x\).

Solution Since \(\sqrt{5x+1}\) appears, we are tempted to set \(u = 5x + 1\). Then \[ \mathrm{d}u = 5\, \mathrm{d}x \implies \sqrt{5x+1}\, \mathrm{d}x = \dfrac15 u^{1/2}\, \mathrm{d}u \] Unfortunately, the integrand is not \(\sqrt{5x+1}\) but \(x\sqrt{5x+1}\). To take care of the extra factor of \(x\), we solve \(u = 5x + 1\) for \(x\) to obtain \(x = \tfrac15 (u-1)\). Then \[ \begin{aligned} x \sqrt{5x+1} \, \mathrm{d}x & = \left(\dfrac15(u-1)\right)\dfrac15 u^{1/2}\, \mathrm{d}u = \dfrac1{25} (u-1)u^{1/2}\, \mathrm{d}u\\ \int x \sqrt{5x+1}\, \mathrm{d}x & =\dfrac1{25} \int (u-1)u^{1/2}\, \mathrm{d}u = \dfrac1{25} \int (u^{3/2} -u^{1/2})\, \mathrm{d}u\\ & = \dfrac1{25}\left(\dfrac25u^{5/2} - \dfrac23u^{3/2}\right) + C\\ & = \dfrac2{125} (5x+1)^{5/2} - \dfrac2{75}(5x+1)^{3/2}+C \end{aligned} \]

Change of Variables Formula for Definite Integrals

The new limits of integration with respect to the \(u\)-variable are \(u(a)\) and \(u(b)\). Think of it this way: As \(x\) varies from \(a\) to \(b\), the variable \(u = u(x)\) varies from \(u(a)\) to \(u(b)\).

Proof

If \(F(x)\) is an antiderivative of \(f(x)\), then \(F(u(x))\) is an antiderivative of \(f(u(x))u'(x)\). FTC I shows that the two integrals are equal: \[ \begin{aligned} \int\limits_a^b f(u(x))u'(x) \, \mathrm{d}x & = F(u(b)) - F(u(a))\\ \int\limits_{u(a)}^{u(b)} f(u) \, \mathrm{d}u & = F(u(b)) - F(u(a)) \end{aligned} \]

Change of Variables for definite integrals: \[ \int\limits_a^b f(u(x)) u'(x) \, \mathrm{d}x = \int\limits_{u(a)}^{u(b)} f(u) \, \mathrm{d}u \]

EXAMPLE 8

Evaluate \(\displaystyle \int\limits_0^2 x^2\sqrt{x^3 + 1}\, \mathrm{d}x\).

Solution Use the substitution \(u = x^{3} + 1\), \(\mathrm{d}u = 3x^{2} \, \mathrm{d}x\): \[ x^2\sqrt{x^3 + 1}\, \mathrm{d}x = \dfrac13\sqrt{u} \, \mathrm{d}u \] By Eq. (2), the new limits of integration \[ u(0) = 0^3 + 1 =1 \quad\text{and}\quad u(2) =2^3 +1 = 9 \] Thus, \[ \int\limits_0^2 x^2\sqrt{x^3 + 1}\, \mathrm{d}x = \dfrac13 \int\limits_1^9 \sqrt{u}\, \mathrm{d}u =\dfrac29u^{3/2}\bigg|_1^9 = \dfrac{52}9 \] This substitution shows that the area in Figure 5.45 is equal to one-third of the area in Figure 5.46 (but note that the figures are drawn to different scales).

Figure 5.45: Region represented by \(\int_0^2 x^2\sqrt{x^3 + 1}\, \mathrm{d}x\).

Figure 5.46: Region represented by \(\int_1^9 \sqrt{u} \, \mathrm{d}u\).

EXAMPLE 9

Evaluate \(\displaystyle \int\limits_0^{\pi/4}\tan^3\theta\sec^2\theta\, \mathrm{d}\theta\).

Solution The substitution \(u = \tan \theta \) makes sense because \(\mathrm{d}u = \sec^{2} \theta \, \mathrm{d}\theta\) and therefore, \(u^{3} \, \mathrm{d}u = \tan^{3} \theta \sec^{2} \theta \, \mathrm{d}\theta\). The new limits of integration are \[ u(0) = \tan 0 = 0 \quad\text{and}\quad u\left(\dfrac\pi4\right)=\tan\left(\dfrac\pi4\right) = 1 \] Thus, \[ \int\limits_0^{\pi/4}\tan^3\theta\sec^2\theta\, \mathrm{d}\theta = \int\limits_0^1u^3 \, \mathrm{d}u = \dfrac{u^4}4\bigg|_0^1 = \dfrac14 \]

EXAMPLE 10

Calculate the area under the graph of \(\displaystyle y = \dfrac{x}{x^2+1}\) over \([1, 3]\).

Solution The area (Figure 5.47) is equal to \(\displaystyle \int\limits_1^3 \dfrac{x}{x^2+1}\, \mathrm{d}x\). We use the substitution

Figure 5.47: Area under the graph of \(\displaystyle y = \dfrac{x}{x^2+1}\) over \([1, 3]\).

The new limits of integration are \(u(1) = 1^{2} + 1 = 2\) and \(u(3) = 3^{2} + 1 = 10\), so \[ \int\limits_1^3 \dfrac{x}{x^2+1}\, \mathrm{d}x = \dfrac12 \int\limits_2^{10} \dfrac{\mathrm{d}u}u = \dfrac12 \ln|u|\bigg|_2^{10} = \dfrac12\ln 10 - \dfrac12\ln 2 \approx 0.805 \]