Chapter 4 · Antiderivatives and Integration

4.3 Integration by Substitution

The chain rule run backwards — spot an inner function and its derivative, rename them $u$ and $du$, and the integral simplifies.

The Fundamental Theorem lets us evaluate integrals — provided we can find an antiderivative, which is not always easy. Integration by substitution finds antiderivatives when the integrand is the result of a chain-rule derivative. It is mainly a visual task: the integrand shows you what to do once you recognise its form.

We look for an integrand of the form $$\int f\big(g(x)\big)\,g^\prime(x)\,dx.$$ For example, in $\displaystyle\int\left(x^2-3\right)^3\,2x\,dx$ we have $f(x)=x^3$, $g(x)=x^2-3$ and $g^\prime(x)=2x$, so the integrand is exactly $f\big(g(x)\big)g^\prime(x)$. We substitute part of the integrand with $u$ and part with $du$ — hence the name, also called change of variables.

📐 Theorem 4.6 — Substitution with indefinite integrals
Let $u=g(x)$, where $g^\prime(x)$ is continuous over an interval, let $f$ be continuous over the corresponding range of $g$, and let $F$ be an antiderivative of $f$. Then $$\int f\big(g(x)\big)g^\prime(x)\,dx=\int f(u)\,du=F(u)+C=F\big(g(x)\big)+C.$$ This is the chain rule read backwards: $\dfrac{d}{dx}F\big(g(x)\big)=f\big(g(x)\big)g^\prime(x)$.
🧭 Problem-Solving Strategy — Integration by substitution
  1. Choose an expression $g(x)$ inside the integrand to call $u$ — ideally one whose derivative $g^\prime(x)$ also appears (up to a constant).
  2. Substitute $u=g(x)$ and $du=g^\prime(x)\,dx$.
  3. The integral should now be in terms of $u$ only. If it cannot be evaluated, go back and choose a different $u$.
  4. Evaluate the integral in $u$.
  5. Write the result in terms of $x$.
✏️ Example 1 — Using substitution to find an antiderivative
Find $\displaystyle\int6x\left(3x^2+4\right)^4dx$.
Step 1 Choose $u=3x^2+4$; then $du=6x\,dx$, which is already in the integrand.
Step 2 $\displaystyle\int6x\left(3x^2+4\right)^4dx=\int u^4\,du=\frac{u^5}{5}+C=\frac{\left(3x^2+4\right)^5}{5}+C$.
Check $\dfrac{d}{dx}\left[\tfrac15\left(3x^2+4\right)^5+1\right]=\left(3x^2+4\right)^4\cdot6x$ — exactly the integrand.
✏️ Example 2 — Using substitution with alteration
Find $\displaystyle\int z\sqrt{z^2-5}\,dz$.
Step 1 Rewrite as $\displaystyle\int z\left(z^2-5\right)^{1/2}dz$ and let $u=z^2-5$, $du=2z\,dz$.
Step 2 The integrand has only $z\,dz$, so multiply the $du$ equation by $\tfrac12$: $\tfrac12du=z\,dz$. Then $$\int z\left(z^2-5\right)^{1/2}dz=\frac12\int u^{1/2}du.$$
Step 3 $\dfrac12\cdot\dfrac{u^{3/2}}{3/2}+C=\dfrac13u^{3/2}+C=\dfrac13\left(z^2-5\right)^{3/2}+C$.
Rule Constants can be fixed this way; variables cannot. If $du$ is off by a factor containing $x$, choose a different $u$.
✏️ Example 3 — Substitution with trigonometric functions
Evaluate $\displaystyle\int\frac{\sin t}{\cos^3t}\,dt$.
Step 1 Let $u=\cos t$, so $du=-\sin t\,dt$. Then $\displaystyle\int\frac{\sin t}{\cos^3t}dt=-\int\frac{du}{u^3}=-\int u^{-3}du$.
Step 2 $-\left(-\tfrac12\right)u^{-2}+C=\dfrac{1}{2u^2}+C=\dfrac{1}{2\cos^2t}+C$.
✏️ Example 4 — Substitution with an exponential function
Evaluate $\displaystyle\int3x^2e^{2x^3}dx$.
Step 1 Let $u=2x^3$ (the exponent), $du=6x^2dx$. The integrand has $3x^2dx=\tfrac12du$.
Step 2 $\displaystyle\int3x^2e^{2x^3}dx=\frac12\int e^u\,du=\frac12e^u+C=\frac12e^{2x^3}+C$.
Interactive 1 — Substitution Moves the Area, Not Its Size

For a definite integral, $\displaystyle\int_a^b f\big(g(x)\big)g^\prime(x)\,dx=\int_{g(a)}^{g(b)}f(u)\,du$. The left region lives on the $x$-axis, the right one on the $u$-axis. Their shapes differ, but the areas always match — watch both totals grow together.

in $x$: $f(g(x))\cdot g'(x)$
in $u$: $f(u)$
$x$—
$u = g(x)$—
area in $x$ so far—
area in $u$ so far—

Exercises

I
Find the antiderivative using the indicated substitution.
  • 1 $\displaystyle\int(x+1)^4dx$; $u=x+1$
  • 2 $\displaystyle\int(x-1)^5dx$; $u=x-1$
  • 3 $\displaystyle\int(2x-3)^{-7}dx$; $u=2x-3$
  • 4 $\displaystyle\int(3x-2)^{-11}dx$; $u=3x-2$
  • 5 $\displaystyle\int x\sqrt{x^2+1}\,dx$; $u=x^2+1$
  • 6 $\displaystyle\int x\sqrt{1-x^2}\,dx$; $u=1-x^2$
  • 7 $\displaystyle\int(x-1)\left(x^2-2x\right)^3dx$; $u=x^2-2x$
  • 8 $\displaystyle\int\left(x^2-2x\right)\left(x^3-3x^2\right)^2dx$; $u=x^3-3x^2$
  • 9 $\displaystyle\int\cos^3\theta\,d\theta$; $u=\sin\theta$ (hint: $\cos^2\theta=1-\sin^2\theta$)
  • 10 $\displaystyle\int\sin^3\theta\,d\theta$; $u=\cos\theta$ (hint: $\sin^2\theta=1-\cos^2\theta$)
Check: the derivative of the answer must retrace the integrand
$f$ with its tangent
$f'$ being traced
$x$—
$f(x)$—
$f'(x)$—
$f$ is…—
1 $du=dx$: $\displaystyle\int u^4du=\frac{(x+1)^5}{5}+C$.
2 $\dfrac{(x-1)^6}{6}+C$.
3 $du=2dx$: $\dfrac12\displaystyle\int u^{-7}du=-\frac{1}{12}(2x-3)^{-6}+C$.
4 $du=3dx$: $\dfrac13\displaystyle\int u^{-11}du=-\frac{1}{30}(3x-2)^{-10}+C$.
5 $du=2x\,dx$: $\dfrac12\displaystyle\int u^{1/2}du=\frac13\left(x^2+1\right)^{3/2}+C$.
6 $du=-2x\,dx$: $-\dfrac12\displaystyle\int u^{1/2}du=-\frac13\left(1-x^2\right)^{3/2}+C$.
7 $du=(2x-2)dx=2(x-1)dx$: $\dfrac12\displaystyle\int u^3du=\frac18\left(x^2-2x\right)^4+C$.
8 $du=\left(3x^2-6x\right)dx=3\left(x^2-2x\right)dx$: $\dfrac13\displaystyle\int u^2du=\frac19\left(x^3-3x^2\right)^3+C$.
9 $\cos^3\theta=\left(1-\sin^2\theta\right)\cos\theta$; $u=\sin\theta$, $du=\cos\theta\,d\theta$: $\displaystyle\int\left(1-u^2\right)du=\sin\theta-\frac{\sin^3\theta}{3}+C$.
10 $\sin^3\theta=\left(1-\cos^2\theta\right)\sin\theta$; $u=\cos\theta$, $du=-\sin\theta\,d\theta$: $-\displaystyle\int\left(1-u^2\right)du=-\cos\theta+\frac{\cos^3\theta}{3}+C$.
II
Use a suitable change of variables to determine the indefinite integral.
  • 1 $\displaystyle\int e^{2x}dx$
  • 2 $\displaystyle\int e^{-3x}dx$
  • d $\displaystyle\int2^x\,dx$
  • 3 $\displaystyle\int3^{-x}dx$
  • 4 $\displaystyle\int x(1-x)^{99}dx$
  • 5 $\displaystyle\int\frac{x^2}{\left(x^3-3\right)^2}dx$
  • 6 $\displaystyle\int t\left(1-t^2\right)^{10}dt$
  • 7 $\displaystyle\int(11x-7)^{-3}dx$
  • 8 $\displaystyle\int x^3\sqrt{1-x^2}\,dx$
  • 9 $\displaystyle\int(7x-11)^4dx$
  • 10 $\displaystyle\int y^5\left(1-y^3\right)^{3/2}dy$
  • 11 $\displaystyle\int\cos^3\theta\sin\theta\,d\theta$
  • 12 $\displaystyle\int\sin^7\theta\cos\theta\,d\theta$
  • 13 $\displaystyle\int\cos\theta(1-\cos\theta)^{99}\sin\theta\,d\theta$
  • 14 $\displaystyle\int\cos^2(\pi t)\sin(\pi t)\,dt$
  • 15 $\displaystyle\int\left(1-\cos^3\theta\right)^{10}\cos^2\theta\sin\theta\,d\theta$
  • 16 $\displaystyle\int2x^3e^{x^4}dx$
  • 17 $\displaystyle\int x^2e^{-x^3}dx$
  • 18 $\displaystyle\int e^x\sqrt{1+e^x}\,dx$
  • 19 $\displaystyle\int e^{\sin x}\cos x\,dx$
  • 20 $\displaystyle\int\sin^2x\cos^3x\,dx$ (hint: $\sin^2x+\cos^2x=1$)
  • 21 $\displaystyle\int e^{\tan x}\sec^2x\,dx$
  • 22 $\displaystyle\int\frac{e^{\ln x}}{x}dx$
  • 23 $\displaystyle\int t\sin\left(t^2\right)\cos\left(t^2\right)dt$
  • 24 $\displaystyle\int t^2\cos^2\left(t^3\right)\sin\left(t^3\right)dt$
  • 25 $\displaystyle\int\frac{e^{\ln(1-t)}}{1-t}dt$
1 $u=2x$: $\tfrac12e^{2x}+C$.
2 $u=-3x$: $-\tfrac13e^{-3x}+C$.
d $\dfrac{2^x}{\ln2}+C$.
3 $u=-x$: $-\dfrac{3^{-x}}{\ln3}+C$.
4 $u=1-x$, $x=1-u$, $dx=-du$: $-\displaystyle\int(1-u)u^{99}du=\frac{(1-x)^{101}}{101}-\frac{(1-x)^{100}}{100}+C$.
5 $u=x^3-3$, $du=3x^2dx$: $\dfrac13\displaystyle\int u^{-2}du=-\frac{1}{3\left(x^3-3\right)}+C$.
6 $u=1-t^2$, $du=-2t\,dt$: $-\dfrac{\left(1-t^2\right)^{11}}{22}+C$.
7 $u=11x-7$: $-\dfrac{1}{22}(11x-7)^{-2}+C$.
8 $u=1-x^2$, $x^2=1-u$, $x\,dx=-\tfrac12du$: $-\tfrac12\displaystyle\int(1-u)u^{1/2}du=\frac{\left(1-x^2\right)^{5/2}}{5}-\frac{\left(1-x^2\right)^{3/2}}{3}+C$.
9 $\dfrac{(7x-11)^5}{35}+C$.
10 $u=1-y^3$, $y^3=1-u$, $y^2dy=-\tfrac13du$: $-\tfrac13\displaystyle\int(1-u)u^{3/2}du=-\frac{2}{15}\left(1-y^3\right)^{5/2}+\frac{2}{21}\left(1-y^3\right)^{7/2}+C$.
11 $u=\cos\theta$: $-\dfrac{\cos^4\theta}{4}+C$.
12 $u=\sin\theta$: $\dfrac{\sin^8\theta}{8}+C$.
13 $u=1-\cos\theta$, $du=\sin\theta\,d\theta$, $\cos\theta=1-u$: $\displaystyle\int(1-u)u^{99}du=\frac{(1-\cos\theta)^{100}}{100}-\frac{(1-\cos\theta)^{101}}{101}+C$.
14 $u=\cos(\pi t)$, $du=-\pi\sin(\pi t)dt$: $-\dfrac{\cos^3(\pi t)}{3\pi}+C$.
15 $u=1-\cos^3\theta$, $du=3\cos^2\theta\sin\theta\,d\theta$: $\dfrac{\left(1-\cos^3\theta\right)^{11}}{33}+C$.
16 $u=x^4$, $du=4x^3dx$: $\tfrac12e^{x^4}+C$.
17 $u=-x^3$: $-\tfrac13e^{-x^3}+C$.
18 $u=1+e^x$: $\tfrac23\left(1+e^x\right)^{3/2}+C$.
19 $u=\sin x$: $e^{\sin x}+C$.
20 $\sin^2x\cos^3x=\sin^2x\left(1-\sin^2x\right)\cos x$; $u=\sin x$: $\dfrac{\sin^3x}{3}-\dfrac{\sin^5x}{5}+C$.
21 $u=\tan x$: $e^{\tan x}+C$.
22 $e^{\ln x}=x$, so the integrand is $1$: $x+C$.
23 $u=\sin\left(t^2\right)$, $du=2t\cos\left(t^2\right)dt$: $\dfrac{\sin^2\left(t^2\right)}{4}+C$.
24 $u=\cos\left(t^3\right)$, $du=-3t^2\sin\left(t^3\right)dt$: $-\dfrac{\cos^3\left(t^3\right)}{9}+C$.
25 $e^{\ln(1-t)}=1-t$, so the integrand is $1$: $t+C$ (for $t<1$).
III
Choose a substitution $u$ and evaluate the indefinite integral. Check each answer by differentiating.
  • 1 $\displaystyle\int x^2\left(x^3+1\right)^5dx$
  • 2 $\displaystyle\int\frac{x}{\sqrt{x^2+9}}\,dx$
  • 3 $\displaystyle\int x\cos\left(x^2\right)dx$
  • 4 $\displaystyle\int\tan x\sec^2x\,dx$
  • 5 $\displaystyle\int\frac{e^{\sqrt x}}{\sqrt x}\,dx$
  • 6 $\displaystyle\int\frac{x}{x^2+4}\,dx$
  • 7 $\displaystyle\int\frac{(\ln x)^2}{x}\,dx$
  • 8 $\displaystyle\int\frac{dx}{1+4x^2}$
  • 9 $\displaystyle\int\frac{e^x}{1+e^{2x}}\,dx$
  • 10 $\displaystyle\int\frac{dx}{\sqrt{9-x^2}}$
Check: differentiate your answer — the right panel must retrace the integrand
$f$ with its tangent
$f'$ being traced
$x$—
$f(x)$—
$f'(x)$—
$f$ is…—
1 $u=x^3+1$, $du=3x^2dx$, so $x^2dx=\tfrac13du$: $\dfrac13\displaystyle\int u^5du=\frac{u^6}{18}+C=\frac{\left(x^3+1\right)^6}{18}+C$.
2 $u=x^2+9$, $du=2x\,dx$: $\dfrac12\displaystyle\int u^{-1/2}du=\frac12\cdot2u^{1/2}+C=\sqrt{x^2+9}+C$.
3 $u=x^2$, $du=2x\,dx$: $\dfrac12\displaystyle\int\cos u\,du=\frac12\sin u+C=\frac12\sin\left(x^2\right)+C$.
4 $u=\tan x$, $du=\sec^2x\,dx$: $\displaystyle\int u\,du=\frac{u^2}{2}+C=\frac{\tan^2x}{2}+C$.
5 $u=\sqrt x$, $du=\dfrac{1}{2\sqrt x}dx$, so $\dfrac{dx}{\sqrt x}=2\,du$: $2\displaystyle\int e^u\,du=2e^u+C=2e^{\sqrt x}+C$.
6 $u=x^2+4$, $du=2x\,dx$: $\dfrac12\displaystyle\int\frac{du}{u}=\frac12\ln|u|+C=\frac12\ln\left(x^2+4\right)+C$ (no absolute value needed since $x^2+4>0$).
7 $u=\ln x$, $du=\dfrac1x\,dx$: $\displaystyle\int u^2du=\frac{u^3}{3}+C=\frac{(\ln x)^3}{3}+C$.
8 Write $1+4x^2=1+(2x)^2$; $u=2x$, $du=2\,dx$: $\dfrac12\displaystyle\int\frac{du}{1+u^2}=\frac12\tan^{-1}u+C=\frac12\tan^{-1}(2x)+C$.
9 $e^{2x}=\left(e^x\right)^2$; $u=e^x$, $du=e^x\,dx$: $\displaystyle\int\frac{du}{1+u^2}=\tan^{-1}u+C=\tan^{-1}\left(e^x\right)+C$.
10 Factor out the $9$: $\sqrt{9-x^2}=3\sqrt{1-(x/3)^2}$. With $u=\tfrac x3$, $dx=3\,du$: $\displaystyle\int\frac{3\,du}{3\sqrt{1-u^2}}=\sin^{-1}u+C=\sin^{-1}\frac x3+C$ (for $|x|<3$).
IV
Definite integrals, back-substitution and applications. In parts 1–5 change the limits of integration to $u$-values rather than returning to $x$.
  • 1 $\displaystyle\int_0^1x\left(x^2+1\right)^3dx$
  • 2 $\displaystyle\int_0^{\pi/2}\sin x\cos^2x\,dx$
  • 3 $\displaystyle\int_0^1xe^{-x^2}dx$
  • 4 $\displaystyle\int_0^4\frac{dx}{\sqrt{2x+1}}$
  • 5 $\displaystyle\int_1^{e^2}\frac{dx}{x\,(1+\ln x)}$
  • 6 $\displaystyle\int x\sqrt{x+1}\,dx$ (hint: let $u=x+1$ and solve for $x$)
  • 7 $\displaystyle\int_0^3x\sqrt{x+1}\,dx$
  • 8 $\displaystyle\int\frac{x}{\sqrt{x+4}}\,dx$
  • 9 Conceptual. With $u=x^2$, both limits of $\displaystyle\int_{-1}^{1}xe^{x^2}dx$ become $u=1$. What is the value of the integral, and why is it consistent with the graph of $xe^{x^2}$?
  • 10 Applied. A particle moves along a line with velocity $v(t)=t\sqrt{t^2+16}$ m/s. How far does it travel during the first $3$ seconds?
1 $u=x^2+1$, $du=2x\,dx$; $x=0\Rightarrow u=1$, $x=1\Rightarrow u=2$: $\dfrac12\displaystyle\int_1^2u^3du=\frac18u^4\Big|_1^2=\frac{16-1}{8}=\frac{15}{8}$.
2 $u=\cos x$, $du=-\sin x\,dx$; $x=0\Rightarrow u=1$, $x=\tfrac\pi2\Rightarrow u=0$: $-\displaystyle\int_1^0u^2du=\int_0^1u^2du=\frac13$.
3 $u=-x^2$, $du=-2x\,dx$; $u$ runs from $0$ to $-1$: $-\dfrac12\displaystyle\int_0^{-1}e^u\,du=\frac12\int_{-1}^0e^u\,du=\frac12\left(1-e^{-1}\right)\approx0.316$.
4 $u=2x+1$, $du=2\,dx$; $u$ runs from $1$ to $9$: $\dfrac12\displaystyle\int_1^9u^{-1/2}du=\sqrt u\,\Big|_1^9=3-1=2$.
5 $u=1+\ln x$, $du=\dfrac1x\,dx$; $x=1\Rightarrow u=1$, $x=e^2\Rightarrow u=3$: $\displaystyle\int_1^3\frac{du}{u}=\ln u\,\Big|_1^3=\ln3\approx1.099$.
6 $u=x+1$ gives $du=dx$ and $x=u-1$: $$\int x\sqrt{x+1}\,dx=\int(u-1)u^{1/2}du=\int\left(u^{3/2}-u^{1/2}\right)du=\frac25(x+1)^{5/2}-\frac23(x+1)^{3/2}+C.$$
7 Same substitution; $u$ runs from $1$ to $4$: $\left[\tfrac25u^{5/2}-\tfrac23u^{3/2}\right]_1^4=\left(\tfrac{64}{5}-\tfrac{16}{3}\right)-\left(\tfrac25-\tfrac23\right)=\tfrac{62}{5}-\tfrac{14}{3}=\dfrac{116}{15}\approx7.733$.
8 $u=x+4$, $x=u-4$, $dx=du$: $\displaystyle\int\frac{u-4}{u^{1/2}}du=\int\left(u^{1/2}-4u^{-1/2}\right)du=\frac23(x+4)^{3/2}-8\sqrt{x+4}+C$.
9 $\displaystyle\int_1^1\tfrac12e^u\,du=0$. This agrees with the graph: $xe^{x^2}$ is odd, so the area below the axis on $[-1,0]$ exactly cancels the area above it on $[0,1]$. Equal $u$-limits signal that $u$ retraces its values and the signed contributions cancel.
10 $v(t)\ge0$ on $[0,3]$, so distance $=\displaystyle\int_0^3t\sqrt{t^2+16}\,dt$. With $u=t^2+16$ ($u$ from $16$ to $25$): $\dfrac12\displaystyle\int_{16}^{25}u^{1/2}du=\frac13u^{3/2}\Big|_{16}^{25}=\frac{125-64}{3}=\frac{61}{3}\approx20.33$ m.
🔬 Research–Teaching NexusResearch-led · optional, about 30 minutes

Research question. How much total drug exposure does the model $C(t)=a\,t\,e^{-bt^2}$ give, exactly and numerically?

Why it matters. Pharmacologists measure the area under the concentration curve (AUC) to decide doses and compare medicines; it is a key number in drug studies.

What to do.

  1. Use sample values $a=20$ mg/L per hour and $b=0.5$ h$^{-2}$. Sketch $C(t)$ in Desmos.
  2. Using the substitution $u=bt^2$, show that $\int_0^T a\,t\,e^{-bt^2}dt=\frac{a}{2b}\left(1-e^{-bT^2}\right)$.
  3. Compute the AUC for $T=4$ h and the limit as $T\to\infty$, which is $\frac{a}{2b}$.
  4. Estimate the same AUC with a trapezoid sum using readings every 0.5 h. Compare.

Tools: Desmos, spreadsheet

Hand in: A half-page note: the exact and numerical AUC, the peak time, and one limitation of the model.

Going further. Anywhere: find how AUC is used in one published drug study summary.

Instructor note: With the sample values, the AUC is about $20$ mg·h/L, and the peak is at $t=\frac{1}{\sqrt{2b}}=1$ h.

🧪 Self-check quiz

0/5 correct

Q1. For $\displaystyle\int6x\left(3x^2+4\right)^4dx$, the natural choice is $u=$

Q2. With $u=z^2-5$, $z\,dz$ equals:

Q3. $\displaystyle\int e^{2x}dx=$

Q4. Substitution is the reverse of which rule?

Q5. $\displaystyle\int\sin^7\theta\cos\theta\,d\theta=$

← 4.2 The Definite Integral4.4 Integration by Parts →