Integration Techniques I: Substitution and Integration by Parts

Chapter 14 reduced the entire problem of computing a definite integral to a single instruction: find an antiderivative, then subtract its values at the endpoints. That reduction is a miracle — it collapses an infinite limit of Riemann sums into one subtraction. But it has a catch. It only helps if you can actually find the antiderivative.

The basic antiderivative table from Chapter 12 — powers, exponentials, sines, cosines, logarithms — handles only the simplest cases. The moment a function is built from those pieces by composition ($\cos(x^2)$), multiplication ($x e^x$), or division ($\frac{x}{x^2+1}$), it leaves the table behind. Differentiation has rules for all three operations: the chain rule, the product rule, the quotient rule. Integration, run in reverse, needs techniques to match.

This chapter develops the two most important ones:

• $u$-substitution — reverses the chain rule. Use it when the integrand is a composite function multiplied by (a constant times) the derivative of the inner function.
• Integration by parts — reverses the product rule. Use it when the integrand is a product of two different kinds of functions (polynomial × exponential, logarithm × algebraic, and so on).

Together they handle a very large fraction of the integrals that arise in scientific work. The remaining specialized techniques — trigonometric integrals, trigonometric substitution, partial fractions, and numerical methods — wait in Chapter 16.

The Key Insight. Every integration technique is a differentiation rule read backwards. $u$-substitution is the chain rule in reverse; integration by parts is the product rule in reverse. You already know the forward rules cold from Chapter 7. Learning to integrate is learning to recognize, in an unfamiliar integrand, the fingerprint that a forward rule would have left behind.

This is also the most-practiced chapter in Part III, and there is a reason. Integration is fundamentally a pattern-recognition skill: you look at an expression and the structure tells you "$u$-substitution" or "integration by parts" before you write a single symbol. That recognition develops only through doing many problems — which is exactly the theme that hand computation builds the understanding that machine computation later amplifies (Section 15.9). Do the exercises. There is no shortcut.

15.1 The Idea Behind $u$-Substitution

Recall the chain rule from Chapter 7: if $F(x) = f(g(x))$, then

$$F'(x) = f'(g(x)) \cdot g'(x).$$

Read that as an antiderivative statement and it becomes

$$\int f'(g(x)) \cdot g'(x) \, dx = f(g(x)) + C.$$

The left-hand side looks forbidding: an outer function's derivative, composed with an inner function, multiplied by the inner function's derivative. The right-hand side is almost trivial — just the outer function evaluated at $g(x)$. The whole difficulty is seeing that the messy left side is secretly the clean right side.

The substitution rule is the bookkeeping device that makes the secret visible. Set $u = g(x)$. Then the differential is $du = g'(x)\,dx$, and the integral on the left transforms into

$$\int f'(u) \, du = f(u) + C = f(g(x)) + C.$$

Nothing new happened mathematically — this is the same identity restated. What the substitution buys you is psychological: the variable change "absorbs" the chain-rule structure, leaving an integral so simple it is in the table.

Geometric Intuition. A substitution is a change of horizontal coordinate. Picture the graph of the integrand over the $x$-axis. Letting $u = g(x)$ rescales and reshapes that axis — stretching it where $g$ is changing fast and compressing it where $g$ is changing slowly. The factor $du = g'(x)\,dx$ is exactly the local stretch factor that keeps the area (the thing the integral measures) unchanged under the warp. A good substitution warps a lumpy region into one with a flat, easy-to-measure profile in the new coordinate. This is the one-dimensional ancestor of the Jacobian change of variables you will meet in Chapter 33.

Worked Example 15.1.1 — A canonical case

Evaluate $\displaystyle\int 2x \cos(x^2) \, dx$.

Recognize the pattern. The integrand contains $\cos(x^2)$ — a composite function — multiplied by $2x$, which happens to be exactly the derivative of the inner function $x^2$.

Set up the substitution. Let $u = x^2$. Then $du = 2x \, dx$. The cluster $2x\,dx$ already sitting in the integrand becomes simply $du$.

Substitute and integrate. $$\int 2x \cos(x^2) \, dx = \int \cos(u) \, du = \sin(u) + C.$$

Back-substitute. Replace $u$ with $x^2$: $$\int 2x \cos(x^2) \, dx = \sin(x^2) + C.$$

Verify. Differentiate the answer: $\frac{d}{dx}\big[\sin(x^2) + C\big] = \cos(x^2)\cdot 2x$. ✓ Always verify; it is one differentiation and it catches almost every error.

The Step-by-Step Procedure

1. Recognize that the integrand has the form $f'(g(x)) \cdot g'(x)$ — an "inner" function and its derivative both present, perhaps up to a constant factor.
2. Choose $u = g(x)$ to be the inner function.
3. Compute $du = g'(x) \, dx$.
4. Substitute to rewrite the integral entirely in terms of $u$ and $du$. No $x$ may remain.
5. Integrate the simpler $\int f'(u) \, du$.
6. Back-substitute $u = g(x)$ into the answer.
7. Verify by differentiating.

The Key Insight. The hard part of $u$-substitution is step 1: recognizing the pattern. Once you see the composite structure with the inner derivative present, steps 2–7 are mechanical. Everything you find difficult about substitution is really difficulty in recognition, and recognition is trained by volume.

When Step 4 Goes Wrong: A Leftover $x$

The single most common breakdown is reaching step 4 and finding an $x$ that refuses to leave. If you cannot eliminate $x$ completely, either the substitution is wrong, or — frequently — you must express the leftover $x$ in terms of $u$.

Example. $\displaystyle\int x \sqrt{x + 1} \, dx$. Let $u = x + 1$, so $du = dx$. But the integrand carries an extra $x$, which is not part of $\sqrt{x+1}$. Solve the substitution backwards: $x = u - 1$. Now everything is in $u$:

$$\int x \sqrt{x+1} \, dx = \int (u - 1) \sqrt{u} \, du = \int \big(u^{3/2} - u^{1/2}\big) \, du = \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2} + C.$$

Back-substitute: $\displaystyle\frac{2}{5}(x+1)^{5/2} - \frac{2}{3}(x+1)^{3/2} + C$.

Check Your Understanding. Evaluate $\displaystyle\int (3x^2 + 2)\,(x^3 + 2x)^5 \, dx$.

Answer

The inner function is $x^3 + 2x$, whose derivative is $3x^2 + 2$ — and there it is, sitting in front. Let $u = x^3 + 2x$, so $du = (3x^2+2)\,dx$. Then $\int u^5\,du = \tfrac{u^6}{6} + C = \tfrac{1}{6}(x^3+2x)^6 + C$. Verify by the chain rule: $\tfrac{d}{dx}\tfrac{1}{6}(x^3+2x)^6 = (x^3+2x)^5(3x^2+2)$. ✓

15.2 More Worked Examples of $u$-Substitution

The reach of $u$-substitution comes from how many superficially different patterns it covers. Each example below trains a slightly different recognition cue. Read them as a vocabulary of "fingerprints."

Worked Example 15.2.1 — A $du$ off by a constant

Evaluate $\displaystyle\int x \cos(x^2) \, dx$ — the same integrand as before, but missing the leading $2$.

Let $u = x^2$, so $du = 2x \, dx$, i.e. $x \, dx = \tfrac{1}{2} du$. The integrand becomes $\cos(u)\cdot\tfrac12\,du$:

$$\int x \cos(x^2) \, dx = \frac{1}{2} \int \cos(u) \, du = \frac{1}{2} \sin(u) + C = \frac{1}{2} \sin(x^2) + C.$$

Constants pass freely through an integral, so a missing factor of $2$ simply reappears as a factor of $\tfrac12$. You may adjust by a constant, never by a variable.

Common Pitfall. Many students try to "fix" a missing variable factor the same way they fix a missing constant — for instance, writing $\int \cos(x^2)\,dx = \frac{1}{2x}\sin(x^2)$ to compensate for the absent $2x$. This is wrong: $\frac{1}{2x}$ is not a constant, so it cannot be pulled outside the integral. In fact $\int \cos(x^2)\,dx$ has no elementary antiderivative at all (it is the Fresnel integral). Substitution rescues you only when the needed factor differs from what is present by a constant.

Worked Example 15.2.2 — Derivative over function

Evaluate $\displaystyle\int \frac{x}{x^2 + 1} \, dx$.

The derivative of the denominator is $2x$; the numerator $x$ is half of it. Let $u = x^2 + 1$, $du = 2x\,dx$, so $x\,dx = \tfrac{du}{2}$:

$$\int \frac{x}{x^2 + 1}\, dx = \frac{1}{2} \int \frac{1}{u}\, du = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln(x^2 + 1) + C.$$

We dropped the absolute-value bars because $x^2 + 1 > 0$ always.

This deserves to be lifted into a named pattern, because it recurs constantly:

$$\boxed{\int \frac{g'(x)}{g(x)} \, dx = \ln|g(x)| + C.}$$

Whenever the numerator is the derivative of the denominator (up to a constant), the integral is a logarithm. This single observation handles a large share of the "rational function" integrals you will meet before partial fractions arrive in Chapter 16.

Worked Example 15.2.3 — Exponential with a linear inner

Evaluate $\displaystyle\int e^{3x + 5} \, dx$.

Let $u = 3x + 5$, $du = 3\,dx$, $dx = \tfrac{du}{3}$:

$$\int e^{3x+5}\, dx = \frac{1}{3} \int e^u \, du = \frac{1}{3} e^u + C = \frac{1}{3} e^{3x+5} + C.$$

For any linear inner function $u = ax + b$, substitution always plays out the same way: the result is $\tfrac{1}{a}$ times "what you would get if the inner were just $x$." This is worth internalizing as a reflex — $\int \sin(5x)\,dx = -\tfrac15\cos(5x)+C$, $\int e^{-2x}\,dx = -\tfrac12 e^{-2x}+C$, all without writing the substitution out.

Worked Example 15.2.4 — A power of sine times cosine

Evaluate $\displaystyle\int \sin^3 x \cos x \, dx$.

Let $u = \sin x$, $du = \cos x \, dx$:

$$\int \sin^3 x \cos x \, dx = \int u^3 \, du = \frac{u^4}{4} + C = \frac{\sin^4 x}{4} + C.$$

This is the simplest case of a trigonometric integral. Whenever a power of $\sin$ multiplies a single $\cos$ (or vice versa), peel off the lone factor to serve as $du$ and substitute. The systematic treatment of $\int \sin^m x\,\cos^n x\,dx$ for general powers is Chapter 16; here it is just a substitution.

Worked Example 15.2.5 — Logarithm as the new variable

Evaluate $\displaystyle\int \frac{1}{x \sqrt{\ln x}} \, dx$.

Let $u = \ln x$, $du = \tfrac{1}{x}\,dx$. The $\tfrac1x\,dx$ in the integrand is exactly $du$:

$$\int \frac{1}{x \sqrt{\ln x}}\, dx = \int \frac{1}{\sqrt{u}}\, du = \int u^{-1/2} \, du = 2 u^{1/2} + C = 2 \sqrt{\ln x} + C.$$

Treating $\ln x$ as the variable eliminated both the logarithm and the awkward $\tfrac1x$ in one stroke. The presence of $\tfrac1x\,dx$ is the cue that $u = \ln x$ wants to be tried.

Worked Example 15.2.6 — The tangent integral

Evaluate $\displaystyle\int \tan x \, dx$.

There is no composite function in sight — until you rewrite $\tan x = \tfrac{\sin x}{\cos x}$ and recognize the "derivative over function" pattern. Let $u = \cos x$, $du = -\sin x \, dx$, so $\sin x \, dx = -du$:

$$\int \tan x \, dx = \int \frac{\sin x}{\cos x}\, dx = -\int \frac{du}{u} = -\ln|u| + C = -\ln|\cos x| + C = \ln|\sec x| + C.$$

This result is worth memorizing: $\displaystyle\int \tan x \, dx = \ln|\sec x| + C$.

Common Pitfall. Many students instinctively try $u = \sin x$ here, because $\sin x$ feels like the "main character." That fails: it gives $du = \cos x\,dx$, but $\cos x$ sits in the denominator, so the substitution leaves the integral as tangled as before. The right choice is $u = \cos x$, whose derivative $-\sin x\,dx$ matches what is in the numerator. The lesson generalizes: the correct $u$ is usually the function whose derivative also appears, even when it is not the most prominent piece.

Worked Example 15.2.7 — A leftover handled by $x^2 = u - 1$

Evaluate $\displaystyle\int \frac{x^3}{\sqrt{x^2 + 1}} \, dx$.

Let $u = x^2 + 1$, $du = 2x\,dx$, so $x^2 = u - 1$. Split off one $x$ to pair with $dx$:

$$\int \frac{x^3}{\sqrt{x^2 + 1}}\, dx = \int \frac{x^2 \cdot x \, dx}{\sqrt{x^2 + 1}} = \int \frac{u - 1}{\sqrt u} \cdot \frac{du}{2} = \frac{1}{2} \int \big(u^{1/2} - u^{-1/2}\big) \, du.$$

Integrate: $\displaystyle\frac{1}{2}\left(\frac{2}{3} u^{3/2} - 2 u^{1/2}\right) + C = \frac{1}{3}(x^2+1)^{3/2} - \sqrt{x^2 + 1} + C$.

Factoring out $\sqrt{x^2+1}$ tidies it: $\displaystyle \frac{1}{3}\sqrt{x^2+1}\,(x^2 - 2) + C$, which is exactly what sympy returns in Section 15.9. The trick, once more, is to rewrite the stubborn extra $x^2$ in terms of $u$.

15.3 Definite Integrals via Substitution

For definite integrals there are two valid approaches, and choosing well saves both time and errors.

Approach 1 — Antiderivative first, then evaluate

Find the antiderivative by substitution, back-substitute all the way to $x$, then apply the original limits.

Example. $\displaystyle\int_0^1 2x e^{x^2} \, dx$.

Step 1, indefinite: let $u = x^2$, $du = 2x\,dx$, giving $\int 2x e^{x^2}\,dx = \int e^u\,du = e^u + C = e^{x^2} + C$.

Step 2, evaluate at the original limits: $$\int_0^1 2x e^{x^2} \, dx = \Big[e^{x^2}\Big]_0^1 = e^{1} - e^{0} = e - 1.$$

Approach 2 — Change the limits

Change the variable and the limits together. If $u = g(x)$, then $x = a$ becomes $u = g(a)$ and $x = b$ becomes $u = g(b)$. You never return to $x$.

Same example. $u = x^2$, $du = 2x\,dx$. When $x = 0$, $u = 0$; when $x = 1$, $u = 1$:

$$\int_0^1 2x e^{x^2} \, dx = \int_0^1 e^u \, du = \Big[e^u\Big]_0^1 = e - 1.$$

Same answer, and cleaner — no back-substitution at all.

Common Pitfall. With Approach 2 you must never evaluate a $u$-antiderivative at the original $x$-limits. The limits must always match the variable currently in the bracket. If your antiderivative is in $u$ and you plug in $x$-values, the answer is generally wrong. The discipline is simple: the moment you change the variable, change the limits in the same breath.

Worked Example 15.3.1 — A substitution that exposes hidden improperness

Evaluate $\displaystyle\int_0^{\pi/4} \tan(2x) \, dx$.

Let $u = 2x$, $du = 2\,dx$, $dx = \tfrac{du}{2}$. When $x = 0$, $u = 0$; when $x = \tfrac{\pi}{4}$, $u = \tfrac{\pi}{2}$:

$$\int_0^{\pi/4} \tan(2x) \, dx = \frac{1}{2} \int_0^{\pi/2} \tan u \, du.$$

But $\tan u$ blows up as $u \to \tfrac{\pi}{2}^-$, so the transformed integral is improper — and therefore the original was improper in disguise. Substitution did not create the trouble; it revealed it. This is a useful warning about Approach 2: always glance at whether the new integrand is well-behaved across the new interval. We treat improper integrals like this one carefully in Chapter 17.

Worked Example 15.3.2 — Symmetry as a shortcut

Evaluate $\displaystyle\int_{-1}^1 \frac{x^3}{1 + x^2} \, dx$.

Before reaching for substitution, look at the structure. The integrand $f(x) = \tfrac{x^3}{1+x^2}$ is odd: $f(-x) = -f(x)$. Over a symmetric interval, the negative area on the left exactly cancels the positive area on the right, so the integral is $0$ with no computation at all.

Geometric Intuition. An odd integrand over $[-a,a]$ encloses a region with half-turn symmetry about the origin: every patch of positive area on the right has a mirror-image patch of negative (below-axis) area on the left. They annihilate. Recognizing parity before you compute can turn a page of algebra into a single glance — a recurring theme of this book, that geometry and algebra are inseparable, and the geometry is often the faster route.

Worked Example 15.3.3 — Change-of-limits done right

Evaluate $\displaystyle\int_1^e \frac{\ln x}{x} \, dx$.

Let $u = \ln x$, $du = \tfrac{dx}{x}$. When $x = 1$, $u = \ln 1 = 0$; when $x = e$, $u = \ln e = 1$:

$$\int_1^e \frac{\ln x}{x}\, dx = \int_0^1 u \, du = \Big[\tfrac{u^2}{2}\Big]_0^1 = \frac{1}{2}.$$

Notice how clean Approach 2 is here: the integrand $\tfrac{\ln x}{x}$ became simply $u$, and the limits $1, e$ became $0, 1$. Had you carelessly kept the $x$-limits and written $\big[\tfrac{u^2}{2}\big]_1^e$, you would be evaluating a $u$-expression at $x$-values — a meaningless mixture that only happens to give the right number here by an accident of $\ln$. Do not rely on accidents; change the limits.

15.4 The Idea Behind Integration by Parts

Substitution reverses the chain rule. The other great technique reverses the product rule. Start from the product rule of Chapter 7, $(fg)' = f'g + fg'$, and integrate both sides:

$$fg = \int f'g \, dx + \int fg' \, dx.$$

Now rearrange to isolate one of the two integrals:

$$\int fg' \, dx = fg - \int f'g \, dx.$$

In the standard streamlined notation we set $u = f$ and $dv = g'\,dx$ (so that $du = f'\,dx$ and $v = g$), and the formula becomes the one you will use thousands of times:

$$\boxed{\int u \, dv = uv - \int v \, du.}$$

This is integration by parts. It does not evaluate an integral directly — it trades one integral, $\int u\,dv$, for another, $\int v\,du$. The entire art lies in choosing $u$ and $dv$ so that the new integral is easier than the one you started with.

Why the Choice Matters

For $\displaystyle\int x e^x \, dx$, you have two options:

• $u = x$, $dv = e^x\,dx$ → $du = dx$, $v = e^x$. New integral: $\int e^x\,dx$ — trivial.
• $u = e^x$, $dv = x\,dx$ → $du = e^x\,dx$, $v = \tfrac{x^2}{2}$. New integral: $\int \tfrac{x^2}{2}e^x\,dx$ — worse than the original (the power of $x$ went up).

The first choice simplifies; the second actively backfires. The difference between them is the whole game.

The LIATE Heuristic

To pick $u$ wisely, run down this priority list and let $u$ be the first type that appears (most preferred first):

• L — Logarithmic ($\ln x$, $\log_b x$)
• I — Inverse trigonometric ($\arcsin x$, $\arctan x$, …)
• A — Algebraic (polynomials, rational functions)
• T — Trigonometric ($\sin x$, $\cos x$, …)
• E — Exponential ($e^x$, $b^x$)

Whatever remains becomes $dv$.

Why this order works. $u$ is the factor that gets differentiated. Logarithms and inverse trig functions become algebraically simpler when differentiated ($\ln x \to \tfrac1x$, $\arctan x \to \tfrac{1}{1+x^2}$) and are miserable to integrate, so we hand them the differentiating role at the top of the list. Polynomials drop their degree on each differentiation, marching toward zero, so they come next. Trig and exponential functions merely cycle under differentiation — neither simpler nor harder — so they are happiest as $dv$, the factor we integrate. LIATE encodes "let the part that gets better when differentiated be $u$."

The Key Insight. Integration by parts moves a "hard piece" from inside an integral to outside it, betting that the leftover integral is easier than the original. The skill is predicting whether the bet pays off before you commit. LIATE is the heuristic that captures the typical winning bet — not a theorem, but right far more often than not. When in doubt, follow it.

LIATE has exceptions, and you will meet a couple. But default to it; it earns its keep.

15.5 Worked Examples of Integration by Parts

Worked Example 15.5.1 — Polynomial × exponential

Evaluate $\displaystyle\int x e^x \, dx$.

LIATE: Algebraic before Exponential, so $u = x$, $dv = e^x\,dx$. Then $du = dx$, $v = e^x$:

$$\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C = (x - 1) e^x + C.$$

Verify. $\tfrac{d}{dx}\big[(x-1)e^x\big] = e^x + (x-1)e^x = x e^x$. ✓

Worked Example 15.5.2 — Polynomial × trig

Evaluate $\displaystyle\int x \cos x \, dx$.

LIATE: Algebraic before Trig, so $u = x$, $dv = \cos x\,dx$. Then $du = dx$, $v = \sin x$:

$$\int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x + C.$$

Worked Example 15.5.3 — A logarithm standing alone

Evaluate $\displaystyle\int \ln x \, dx$.

There is no visible product — but you can manufacture one by writing the integrand as $\ln x \cdot 1$. LIATE makes $\ln$ the $u$: take $u = \ln x$, $dv = 1\cdot dx$. Then $du = \tfrac1x\,dx$, $v = x$:

$$\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x}\, dx = x \ln x - \int 1\,dx = x \ln x - x + C.$$

Memorize this — $\displaystyle\int \ln x \, dx = x \ln x - x + C$ appears everywhere. The "multiply by an invisible $1$" trick recovers every inverse-function antiderivative.

Worked Example 15.5.4 — Inverse trig (parts feeding substitution)

Evaluate $\displaystyle\int \arctan x \, dx$.

Same trick: $u = \arctan x$, $dv = dx$, so $du = \tfrac{1}{1+x^2}\,dx$, $v = x$:

$$\int \arctan x \, dx = x \arctan x - \int \frac{x}{1 + x^2}\, dx.$$

The leftover integral is precisely Worked Example 15.2.2, a $u$-substitution: $\int \tfrac{x}{1+x^2}\,dx = \tfrac12\ln(1+x^2) + C$. Therefore

$$\int \arctan x \, dx = x \arctan x - \frac{1}{2} \ln(1 + x^2) + C.$$

Notice the pattern: integration by parts often hands you a substitution problem as its second act. The two techniques are partners, not rivals.

Worked Example 15.5.5 — Twice by parts

Evaluate $\displaystyle\int x^2 e^x \, dx$.

First pass: $u = x^2$, $dv = e^x\,dx$, so $du = 2x\,dx$, $v = e^x$:

$$\int x^2 e^x \, dx = x^2 e^x - \int 2x e^x \, dx.$$

The leftover integral is twice Worked Example 15.5.1: $\int 2x e^x\,dx = 2(x-1)e^x$. Assembling,

$$\int x^2 e^x \, dx = x^2 e^x - 2(x - 1) e^x + C = (x^2 - 2x + 2) e^x + C.$$

Generalization. $\int x^n e^x\,dx$ needs $n$ passes, the power of $x$ dropping by one each time until it hits zero. That repetition is exactly the kind of mechanical labor the tabular method (below) automates.

Worked Example 15.5.6 — Three times by parts

Evaluate $\displaystyle\int x^3 \sin x \, dx$.

Pass 1: $u = x^3$, $dv = \sin x\,dx$ → $v = -\cos x$, $du = 3x^2\,dx$: $$\int x^3 \sin x \, dx = -x^3 \cos x + \int 3 x^2 \cos x \, dx.$$

Pass 2 on $\int 3x^2 \cos x\,dx$: $u = 3x^2$, $dv = \cos x\,dx$ → $v = \sin x$, $du = 6x\,dx$: $$\int 3 x^2 \cos x \, dx = 3 x^2 \sin x - \int 6 x \sin x \, dx.$$

Pass 3 on $\int 6x \sin x\,dx$: $u = 6x$, $dv = \sin x\,dx$ → $v = -\cos x$, $du = 6\,dx$: $$\int 6 x \sin x \, dx = -6x \cos x + \int 6 \cos x \, dx = -6x \cos x + 6 \sin x.$$

Threading the three together (mind the signs): $$\int x^3 \sin x \, dx = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x + C.$$

Mechanical, correct, and exhausting. The sign-tracking is where errors creep in — which is exactly the problem the tabular method solves.

The Tabular Method

For $\int p(x)\,f(x)\,dx$ where $p$ is a polynomial and $f$ is easy to integrate repeatedly, you can organize all the by-parts passes into one table. Build three columns: alternating signs, repeated derivatives of $p$ (until you reach $0$), and repeated integrals of $f$.

For $\int x^3 \sin x\,dx$:

Now multiply along the diagonals — each row's derivative entry times the next row's integral entry, carrying the sign:

$$+\,x^3(-\cos x)\;-\;3x^2(-\sin x)\;+\;6x(\cos x)\;-\;6(\sin x) = -x^3 \cos x + 3 x^2 \sin x + 6x \cos x - 6 \sin x.$$

Identical to the three-pass result, with a fraction of the writing and almost no chance of a sign slip.

Computational Note. The table terminates because differentiating a degree-$n$ polynomial $n+1$ times yields $0$. For factors that never differentiate to zero (say $e^x$ as the column being differentiated), the table runs forever — so always put the polynomial in the differentiation column. When neither factor terminates (e.g. $e^x \times \sin x$), tabular integration does not close; that case is exactly what the rotating trick in Section 15.6 is for.

Check Your Understanding. Use the tabular method to evaluate $\displaystyle\int x^2 \cos x \, dx$.

Answer

Derivatives of $x^2$: $x^2,\ 2x,\ 2,\ 0$. Integrals of $\cos x$: $\cos x,\ \sin x,\ -\cos x,\ -\sin x$. Signs $+,-,+$. Diagonal products: $+x^2\sin x - 2x(-\cos x) + 2(-\sin x) = x^2\sin x + 2x\cos x - 2\sin x + C$. Differentiate to check: it returns $x^2\cos x$. ✓

15.6 The "Rotating" Trick

Occasionally integration by parts, after a pass or two, regenerates the original integral. Far from a dead end, this is a gift: treat the unknown integral as an algebraic variable and solve for it.

Worked Example 15.6.1 — Exponential × trig

Evaluate $I = \displaystyle\int e^x \sin x \, dx$.

Here LIATE is ambivalent — both factors merely cycle under differentiation — so the choice of $u$ does not matter, as long as you are consistent. Take $u = e^x$, $dv = \sin x\,dx$, so $du = e^x\,dx$, $v = -\cos x$:

$$I = -e^x \cos x + \int e^x \cos x \, dx.$$

Apply parts again to $\int e^x \cos x\,dx$, keeping $u = e^x$ (the same kind of choice): $dv = \cos x\,dx$, $v = \sin x$, $du = e^x\,dx$:

$$\int e^x \cos x \, dx = e^x \sin x - \int e^x \sin x \, dx = e^x \sin x - I.$$

Substitute back into the first line:

$$I = -e^x \cos x + e^x \sin x - I.$$

The original $I$ has reappeared on the right. Solve algebraically — add $I$ to both sides:

$$2 I = e^x (\sin x - \cos x) \qquad\Longrightarrow\qquad I = \frac{e^x (\sin x - \cos x)}{2} + C.$$

This is one of the most elegant maneuvers in all of integration. It generalizes cleanly: any product $e^{ax}\sin(bx)$ or $e^{ax}\cos(bx)$ yields to a two-pass by-parts plus this algebraic solve.

Warning. In the second pass you must keep $u$ as $e^x$. If you instead pick $u = \cos x$, the second integration by parts simply undoes the first, returning you to $\int e^x\sin x\,dx$ and an identity $I = I$ that says nothing. The trick lives or dies on choosing the "same direction" both times. This is a genuine trap, not a slip of arithmetic — it derails the entire method.

Worked Example 15.6.2 — A second rotating instance: $\sec^3 x$

Evaluate $\displaystyle\int \sec^3 x \, dx$.

Split $\sec^3 x = \sec x \cdot \sec^2 x$, and take $u = \sec x$, $dv = \sec^2 x\,dx$, so $du = \sec x \tan x\,dx$, $v = \tan x$:

$$\int \sec^3 x \, dx = \sec x \tan x - \int \tan x \cdot \sec x \tan x \, dx = \sec x \tan x - \int \sec x \tan^2 x \, dx.$$

Use the Pythagorean identity $\tan^2 x = \sec^2 x - 1$:

$$= \sec x \tan x - \int \sec x(\sec^2 x - 1)\, dx = \sec x \tan x - \int \sec^3 x \, dx + \int \sec x \, dx.$$

The original $\int \sec^3 x\,dx$ is back. Move it to the left:

$$2 \int \sec^3 x \, dx = \sec x \tan x + \int \sec x \, dx.$$

Using the known antiderivative $\int \sec x\,dx = \ln|\sec x + \tan x| + C$,

$$\int \sec^3 x \, dx = \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln|\sec x + \tan x| + C.$$

This particular integral surfaces whenever you compute the arc length or surface area of a parabola or a catenary (Chapter 18), so the result is worth keeping handy.

15.7 Strategy: Which Technique First?

Confronted with an unfamiliar integral, run through this checklist roughly in order — it is sorted by how often each technique is the one that works.

1. Is it already in the standard table? ($\int x^n$, $\int \sin x$, $\int e^x$, $\int \tfrac{1}{x}$, …) If so, write the answer and stop.
2. Is there a composite $f(g(x))$ with $g'(x)$ also present (up to a constant)? → $u$-substitution (Sections 15.1–15.3).
3. Is it a product of two different types of functions (polynomial × exponential, log × algebraic, inverse-trig alone)? → integration by parts, choosing $u$ by LIATE (Sections 15.4–15.6).
4. Does it involve products or powers of trig functions, $\sin^m x \cos^n x$? → trigonometric integrals (Chapter 16).
5. Does it contain $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$? → trigonometric substitution (Chapter 16).
6. Is it a rational function $\tfrac{P(x)}{Q(x)}$ with no obvious substitution? → partial fractions (Chapter 16).
7. Does it have no elementary antiderivative (e.g. $\int e^{-x^2}\,dx$, $\int \tfrac{\sin x}{x}\,dx$)? → numerical methods (Chapter 16), or express it through special functions.

Common Pitfall. The most expensive strategic error is committing to integration by parts on something that is really a substitution (or vice versa) and grinding for half a page before noticing. Spend the first ten seconds classifying the integrand, not computing. A composite-with-inner-derivative is substitution; a product-of-different-types is parts. Misclassify and you can burn an entire exam question on an approach that was never going to close.

Real-World Application — Computer algebra and the limits of cleverness (every quantitative field). Working scientists and engineers rarely integrate $\sec^3 x$ by hand twice in a career; they call a computer algebra system — sympy, Mathematica, Maple, Maxima. These tools implement exactly the techniques of this chapter and Chapter 16, plus deeper machinery like Risch's algorithm, a genuine decision procedure that determines whether an elementary antiderivative even exists and produces it when it does. So why learn the techniques by hand? Because recognition is judgment the machine cannot supply for you: knowing which form to expect, when a closed form should exist, and when to abandon symbols for numerics is what separates using the tool from being used by it. This is the theme that hand computation builds understanding while machine computation builds power — you need both.

15.8 Combined Techniques

Real integrals rarely announce a single method. Many require substitution and by-parts, sometimes more than once. The skill is decomposing a hard integral into a sequence of recognizable moves.

Worked Example 15.8.1 — Substitution, then by-parts

Evaluate $\displaystyle\int \cos(\sqrt{x}) \, dx$.

Step 1 — substitution. The composite $\cos(\sqrt x)$ begs for $w = \sqrt{x}$. Then $x = w^2$ and $dx = 2w\,dw$:

$$\int \cos(\sqrt{x}) \, dx = \int \cos(w) \cdot 2w \, dw = 2 \int w \cos w \, dw.$$

Step 2 — by-parts. The result is a polynomial × trig product. Take $u = w$, $dv = \cos w\,dw$, so $v = \sin w$:

$$\int w \cos w \, dw = w \sin w - \int \sin w \, dw = w \sin w + \cos w + C.$$

Combine and back-substitute $w = \sqrt x$:

$$\int \cos(\sqrt x) \, dx = 2\big(w \sin w + \cos w\big) + C = 2\sqrt x \,\sin\sqrt x + 2 \cos\sqrt x + C.$$

The substitution converted an un-integrable-looking composite into a standard by-parts problem — a typical two-stage attack.

Worked Example 15.8.2 — By-parts that spawns a substitution

Evaluate $\displaystyle\int \arctan(\sqrt x) \, dx$.

By parts (LIATE puts the inverse-trig first): $u = \arctan\sqrt x$, $dv = dx$, so $v = x$ and, by the chain rule, $du = \tfrac{1}{1 + x} \cdot \tfrac{1}{2\sqrt x}\,dx$:

$$\int \arctan\sqrt x \, dx = x \arctan\sqrt x - \int \frac{x}{2 \sqrt x (1 + x)}\, dx = x \arctan\sqrt x - \frac{1}{2} \int \frac{\sqrt x}{1 + x}\, dx.$$

The leftover is a substitution. Let $t = \sqrt x$, so $x = t^2$, $dx = 2t\,dt$:

$$\int \frac{\sqrt x}{1 + x}\, dx = \int \frac{t}{1 + t^2} \cdot 2t \, dt = 2 \int \frac{t^2}{1 + t^2}\, dt = 2 \int \left(1 - \frac{1}{1 + t^2}\right) dt = 2t - 2 \arctan t + C.$$

(That middle step — rewriting $\tfrac{t^2}{1+t^2} = 1 - \tfrac{1}{1+t^2}$ — is a polynomial-division trick you will formalize under partial fractions in Chapter 16.) Back-substitute $t = \sqrt x$: the leftover equals $2\sqrt x - 2\arctan\sqrt x$. Assembling everything,

$$\int \arctan\sqrt x \, dx = x \arctan\sqrt x - \frac{1}{2}\big(2\sqrt x - 2 \arctan\sqrt x\big) + C = (x + 1) \arctan\sqrt x - \sqrt x + C,$$

which matches sympy's x*atan(√x) + atan(√x) - √x from Section 15.9. Intricate, but every move was one you have seen.

Worked Example 15.8.3 — A Gaussian moment

Evaluate $\displaystyle\int x^3 e^{-x^2} \, dx$.

Step 1 — substitution. Split off one $x$ for the $du$ and substitute $u = x^2$, $du = 2x\,dx$, with $x^2 = u$:

$$\int x^3 e^{-x^2}\, dx = \int x^2 \cdot x e^{-x^2}\, dx = \frac{1}{2} \int u e^{-u}\, du.$$

Step 2 — by-parts on $\int u e^{-u}\,du$: $w = u$, $dv = e^{-u}\,du$, so $v = -e^{-u}$:

$$\int u e^{-u}\, du = -u e^{-u} + \int e^{-u}\, du = -u e^{-u} - e^{-u} + C = -(u + 1) e^{-u} + C.$$

Back-substitute $u = x^2$:

$$\int x^3 e^{-x^2}\, dx = -\frac{1}{2}(x^2 + 1) e^{-x^2} + C.$$

Integrals of exactly this shape — $\int x^n e^{-x^2}\,dx$ — are the moments of the Gaussian distribution, the workhorses of statistical mechanics and probability. They connect directly to the area-under-the-normal-curve anchor example introduced in Chapter 13: although $\int e^{-x^2}\,dx$ itself has no elementary form (you will tame it with Taylor series in Chapter 23), every integrand with an odd power of $x$ in front, like this one, integrates in closed form because the lone $x$ supplies the $du$ that substitution needs.

15.9 Symbolic Computation with sympy

You should be building genuine fluency in $u$-substitution and integration by parts by hand — that is where the understanding lives. But for verifying your work and for integrals too tedious to trust to a pencil, sympy is invaluable. The pattern is our usual three tiers: pose it analytically, solve it by hand, then let the machine confirm.

# Verify the chapter's worked integrals symbolically with sympy.
import sympy as sp

x = sp.symbols('x')

# Section 15.2 — u-substitution examples
print(sp.integrate(2*x * sp.cos(x**2), x))   # sin(x**2)
print(sp.integrate(x / (x**2 + 1), x))       # log(x**2 + 1)/2
print(sp.integrate(sp.exp(3*x + 5), x))      # exp(3*x + 5)/3
print(sp.integrate(sp.tan(x), x))            # -log(cos(x))   (= log|sec x|)
print(sp.integrate(x**3 / sp.sqrt(x**2+1), x))
#   -> x**2*sqrt(x**2 + 1)/3 - 2*sqrt(x**2 + 1)/3   (= sqrt(x**2+1)*(x**2-2)/3)

# Section 15.5 — integration by parts examples
print(sp.integrate(x * sp.exp(x), x))        # (x - 1)*exp(x)
print(sp.integrate(sp.log(x), x))            # x*log(x) - x
print(sp.integrate(x**2 * sp.exp(x), x))     # (x**2 - 2*x + 2)*exp(x)
print(sp.integrate(sp.atan(x), x))           # x*atan(x) - log(x**2 + 1)/2

# Section 15.6 — the rotating trick
print(sp.integrate(sp.exp(x) * sp.sin(x), x))  # exp(x)*sin(x)/2 - exp(x)*cos(x)/2

# Section 15.8 — combined techniques
print(sp.integrate(sp.cos(sp.sqrt(x)), x))   # 2*sqrt(x)*sin(sqrt(x)) + 2*cos(sqrt(x))
print(sp.integrate(x**3 * sp.exp(-x**2), x)) # (-x**2 - 1)*exp(-x**2)/2

# Definite integrals — sympy handles the change of limits internally
print(sp.integrate(2*x * sp.exp(x**2), (x, 0, 1)))   # E - 1
print(sp.integrate(x * sp.exp(-x), (x, 0, sp.oo)))   # 1

Every line above prints the hand result we derived, sometimes in a differently-grouped but equivalent form (compare $\tfrac13\sqrt{x^2+1}(x^2-2)$ with sympy's factored output). When sympy and your hand answer look different, the reconciliation — usually one algebraic identity — is itself good practice.

Computational Note. sympy writes log for the natural logarithm, where mathematics conventionally writes $\ln$. This convention is nearly universal in scientific computing: Mathematica's Log[x], NumPy's np.log, and C's log() are all natural logs (base $e$), not base $10$. When sympy returns -log(cos(x)) for $\int\tan x\,dx$, read it as $-\ln|\cos x|$ — and note that sympy omits the absolute-value bars and the $+C$, both of which you must restore for a complete antiderivative.

Check Your Understanding. Without running it, predict what sp.integrate(x*sp.cos(x), x) returns, then reconcile it with Worked Example 15.5.2.

Answer

It returns x*sin(x) + cos(x), exactly the antiderivative $x\sin x + \cos x$ we found by parts — minus the $+C$, which sympy always drops for indefinite integrals. Differentiating the output by the product rule recovers $\cos x - x\sin x + \ldots = x\cos x$. ✓

15.10 Applications Across Every Field

The same two techniques recur, in different costumes, throughout the quantitative sciences. Here are five, deliberately spread across biology, statistics, physics, economics, and engineering, to make the point that calculus appears in every quantitative field.

Application 1 — Drug exposure, AUC (pharmacology)

After a dose, a drug's blood concentration often decays exponentially, $C(t) = C_0 e^{-kt}$. The total drug exposure — the area under the concentration curve (AUC) — is

$$\text{AUC} = \int_0^\infty C_0 e^{-kt} \, dt = \left[-\frac{C_0}{k} e^{-kt}\right]_0^\infty = 0 - \left(-\frac{C_0}{k}\right) = \frac{C_0}{k}.$$

A single $u$-substitution ($u = -kt$) or a direct table antiderivative gives a clean closed form. AUC is the central quantity in pharmacokinetics: oral bioavailability $F = \text{AUC}_{\text{oral}} / \text{AUC}_{\text{IV}}$ measures how much of a swallowed dose actually reaches the bloodstream, and regulators require it to prove two formulations are bioequivalent. This is the net-change reading of FTC from Chapter 14, applied to a decaying rate.

Application 2 — The mean of an exponential distribution (statistics)

If waiting times $X$ follow an exponential distribution with density $f(x) = \lambda e^{-\lambda x}$ for $x \ge 0$, the expected wait is

$$E[X] = \int_0^\infty x \cdot \lambda e^{-\lambda x} \, dx.$$

This is integration by parts: $u = x$, $dv = \lambda e^{-\lambda x}\,dx$, so $du = dx$, $v = -e^{-\lambda x}$:

$$E[X] = \Big[-x e^{-\lambda x}\Big]_0^\infty + \int_0^\infty e^{-\lambda x}\, dx = 0 + \frac{1}{\lambda} = \frac{1}{\lambda}.$$

The boundary term vanishes because $x e^{-\lambda x} \to 0$ as $x \to \infty$ — exponential decay outruns linear growth, a fact you will see again in every Laplace transform. So the mean of an exponential is $1/\lambda$, the reciprocal of its rate. If calls arrive at $\lambda = 0.5$ per minute, the mean gap between calls is $1/0.5 = 2$ minutes.

Application 3 — Work done by a varying force (physics)

Hooke's law gives a spring's restoring force as $F(x) = kx$. The work to stretch it from rest to extension $L$ is

$$W = \int_0^L k x \, dx = \frac{k L^2}{2},$$

the elastic potential energy. Stretching twice as far costs four times the work — energy scales as the square of displacement, the signature of a quadratic potential. For an inverse-square force such as gravity, $F(r) = -\tfrac{GMm}{r^2}$, the work moving from $r_1$ to $r_2$ is a $u$-substitution (or direct antiderivative of $r^{-2}$):

$$W = \int_{r_1}^{r_2} \left(-\frac{GMm}{r^2}\right) dr = GMm\left[\frac{1}{r_2} - \frac{1}{r_1}\right],$$

which is exactly the change in gravitational potential energy.

Application 4 — Consumer surplus (economics)

When a market clears at price $p^*$, the consumer surplus — the total value buyers receive above what they pay — can be written as the area to the left of the demand curve above the clearing price:

$$\text{CS} = \int_{p^*}^{p_{\max}} D(p) \, dp,$$

where $D(p)$ is the demand function and $p_{\max}$ is the choke price at which demand vanishes. For exponential demand $D(p) = a e^{-bp}$ with $p_{\max} \to \infty$, this is a direct exponential integral:

$$\text{CS} = \int_{p^*}^{\infty} a e^{-bp}\, dp = \frac{a}{b} e^{-b p^*}.$$

Richer demand curves — linear-times-exponential, or log demand — pull integration by parts into the calculation. Either way, surplus is an accumulated quantity, the integral interpretation of FTC at work in a market.

Application 5 — The Laplace transform (engineering)

The Laplace transform of a function $f(t)$ is

$$F(s) = \int_0^\infty f(t)\, e^{-st} \, dt,$$

and it is among the most-used tools in all of engineering for solving differential equations (you will meet it again in Chapter 19). Computing the transform of basic signals — $t^n$, $\sin(at)$, $\cos(at)$ — runs almost entirely on integration by parts. For example,

$$\mathcal{L}\{t\}(s) = \int_0^\infty t e^{-st}\, dt = \Big[-\tfrac{t}{s} e^{-st}\Big]_0^\infty + \frac{1}{s}\int_0^\infty e^{-st}\, dt = 0 + \frac{1}{s}\cdot\frac{1}{s} = \frac{1}{s^2}.$$

The entry $t \leftrightarrow 1/s^2$ sits in every Laplace transform table — and you just derived it with the technique of Section 15.5.

Historical Note. The phrase integration by parts and the modern $\int u\,dv = uv - \int v\,du$ notation descend directly from Gottfried Leibniz (1646–1716), who built the differential and integral calculus around the product rule for differentials, $d(uv) = u\,dv + v\,du$. Reading that identity backwards is, quite literally, integration by parts — the technique fell out of the notation almost for free. It is one more reason continental mathematicians, armed with Leibniz's symbols, raced ahead of Newton's England for a century, as recounted in Chapter 14.

15.11 A Few Integrals Worth Memorizing

Beyond the basic antiderivative table, these results recur often enough — and are awkward enough to re-derive — that committing them to memory genuinely saves time:

You derived every one of these in this chapter; the table is a recall aid, not a substitute for knowing why each holds.

Add to Your Modeling Portfolio. Locate the rate function at the heart of your modeling project and try to integrate it in closed form using this chapter's two techniques. Biology: integrate a dosing or decay rate such as $C_0 e^{-kt}$ to get total drug exposure (AUC), or a logistic growth rate to recover accumulated population. Economics: integrate a marginal-cost or demand function (e.g. $a e^{-bp}$) to compute total cost or consumer surplus, using by-parts when a polynomial factor appears. Physics: integrate a position-dependent force $F(x)$ along a displacement to obtain work, the way we did for the spring and for gravity. Data Science: integrate a probability density times $x$ (or $x^2$) by parts to obtain the mean (or a moment) of your distribution. In your portfolio, record whether your model admits a clean closed form or whether it forces you into numerical evaluation (Chapter 16) — that distinction is itself a finding worth reporting.

15.12 Summary

You now hold the two most important integration techniques in the subject.

• $u$-substitution reverses the chain rule. Hunt for the structure $f(g(x))\cdot g'(x)$: an inner function with its derivative present, up to a constant. Choose $u = g(x)$, rewrite everything in $u$, integrate, back-substitute, verify.
• Integration by parts reverses the product rule. For a product of two different types of functions, pick $u$ by LIATE — the factor that simplifies when differentiated — and apply $\int u\,dv = uv - \int v\,du$.

And several refinements that turn these from tricks into a toolkit:

• For definite integrals, you may either keep the original limits (back-substitute first) or change the limits along with the variable — just never mix a $u$-antiderivative with $x$-limits.
• Repeated integration by parts handles polynomials times exponentials or trig functions; the tabular method automates the bookkeeping when one factor differentiates to zero.
• The rotating trick solves integrals like $\int e^x\sin x\,dx$ and $\int\sec^3 x\,dx$ algebraically, by recognizing the original integral on the right-hand side and solving for it.
• Real problems often combine substitution and parts in sequence; decompose them one recognizable move at a time.

Together these techniques dispatch the majority of integrals you will meet in scientific work — and where they fail, they fail informatively, telling you which of the Chapter 16 specialized methods (trigonometric integrals, trigonometric substitution, partial fractions, numerical integration) to reach for next.

Looking Ahead

Chapter 16 extends the toolkit to the harder cases: powers and products of trig functions, the radical-clearing substitutions $x = a\sin\theta$ and $x = a\tan\theta$, the partial-fraction decomposition of rational functions, and numerical integration for the integrals — like $\int e^{-x^2}\,dx$ — that have no elementary antiderivative at all. Chapter 17 confronts improper integrals, the kind that quietly appeared in Worked Example 15.3.1 and in three of this chapter's applications, where a limit runs to infinity or the integrand blows up. Chapter 18 turns all of this loose on geometry and physics — areas between curves, volumes of revolution, arc length (where $\int\sec^3 x\,dx$ resurfaces), and work — and Chapter 19 closes Part III by using integration to solve differential equations.

Reflection

Differentiation has rules; integration has techniques. That asymmetry is not a defect — it is the deepest fact about the subject. Every derivative can be computed mechanically by following the chain, product, and quotient rules in sequence. But running those rules backward is genuinely creative work: you must recognize which forward rule left the fingerprint in front of you, and sometimes no elementary rule did. $u$-substitution and integration by parts are your answers to the two most common fingerprints — the chain rule's composite structure and the product rule's mismatched factors. Master them and most integrals you encounter become routine; the rest you will learn to recognize as genuinely hard, which is its own kind of knowledge. The path to that mastery is the oldest advice in mathematics, and it is honest: do many problems. Then turn the page, because Chapter 16 has more.

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