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The chain rule read backwards, choosing the substitution, and what happens to the limits.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to recognise the shape substitution can undo — a composite times the inner function's derivative — choose $u$ and verify that no $x$ survives before committing, carry a definite integral through either by converting the limits or by substituting back, adjust for a missing constant factor and convert a stray factor by solving back for $x$, and say why substitution fails on integrals that have no elementary antiderivative at all.
The chain rule says $\dfrac{d}{dx}F(g(x)) = F'(g(x))\,g'(x)$. Reading it right to left: $$\int F'(g(x))\,g'(x)\,dx = F(g(x)) + C.$$
With $u = g(x)$ and $du = g'(x)\,dx$, that is $$\int F'(u)\,du = F(u) + C,$$ which is the substitution rule. Nothing new is being asserted — it is the chain rule, written in the other direction.
The shape it needs. The chain rule produces a product: a composite times the inner function's derivative. So substitution works when the integrand has that shape — a composite, and the inner derivative as a factor, up to a constant. Containing a composite is not enough: $\int xe^{x^2}dx$ yields and $\int e^{x^2}dx$ does not.
For a definite integral, the limits are values of $x$. Two correct routes:
Doing neither — evaluating a $u$-integral at $x$ limits — is the standard error, and it produces a plausible number with nothing to flag it.
For an indefinite integral you must substitute back. An answer in $u$ answers a question nobody asked.
Another way: picture
A substitution is a change of ruler. The variable $u$ measures the axis differently, stretching it where $g$ moves fast and compressing it where $g$ moves slowly, and $du = g'(x)\,dx$ is the conversion factor between the two rulers. The area is unchanged; only the coordinates it is described in have moved.
Another way: steps
Almost always the inner function of a composite. In order of how often it works:
| Integrand contains | Try |
|---|---|
| a bracket raised to a power | $u = $ the bracket |
| a root | $u = $ the radicand |
| an exponent | $u = $ the exponent |
| a denominator | $u = $ the denominator |
| $\ln$ of something | $u = $ that something, or sometimes $u = \ln(\cdot)$ |
The test is step 3, and it is decisive: rewrite the integrand in $u$ and see whether any $x$ survives.
A useful variant when it nearly works: solve back for $x$. For $\int x\sqrt{x-1}\,dx$ take $u = x - 1$, so $du = dx$ and $x = u + 1$, giving $\int (u+1)\sqrt u\,du = \int (u^{3/2} + u^{1/2})du$. The leftover $x$ was converted rather than cancelled, which is legitimate whenever $g$ is invertible on the interval.
And sometimes the constant is all that is missing: $\int x^2(x^3+1)^5dx$ with $u = x^3+1$ has $du = 3x^2dx$, so $x^2dx = \tfrac13 du$. A constant factor can always be adjusted; a factor of $x$ cannot.
Evaluating $u$-limits as $x$-limits, or the reverse. Decide which route you are taking and carry it through. This is the commonest error in the lesson and the hardest to see afterwards.
Forgetting to substitute back in an indefinite integral. The answer must be a function of $x$.
Forgetting $dx \to du$. $\int (3x+1)^4dx$ is not $\frac{(3x+1)^5}{5}$; the $du = 3dx$ contributes a factor $\frac13$. Differentiating the answer catches it instantly.
Choosing $u$ without checking step 3. If an $x$ remains, the substitution has not worked, and pressing on produces nonsense.
Expecting it always to work. $\int e^{x^2}dx$, $\int \frac{\sin x}{x}dx$ and $\int \sqrt{1+x^4}\,dx$ have no elementary antiderivatives. No substitution will find one, and that is a theorem rather than a gap in technique.
Substituting across a point where $g$ is not monotone, in a definite integral, without care: the $u$ limits can then fail to describe the same region.
It gets reached for whenever an integral looks difficult, and it only works on one shape: a composite multiplied by the inner function's derivative. That is not a restriction someone imposed — it is what the chain rule produces, and substitution is the chain rule reversed, so it can only undo what the chain rule made.
The test in step 3 decides it in one line. Write $du$, look at what the integrand still holds, and if an $x$ survives that no conversion removes, this is not a substitution problem.
Which leads to the honest summary of where this course ends. Differentiation is an algorithm and always terminates. Integration is a collection of techniques — substitution here, then parts, partial fractions and trigonometric substitution in Calculus II — none of which is complete, and some elementary functions are provably beyond all of them. $\int e^{-x^2}dx$ is not waiting for a cleverer method. The asymmetry is real, it is permanent, and recognising which integrals are inside the reach of a technique is as much of the skill as executing one.
$\displaystyle\int_0^2 x(x^2+1)^3dx$ with $u = x^2+1$, $du = 2x\,dx$, so $x\,dx = \tfrac12 du$.
Check: $x\,dx$ is exactly what is left over.
Route 1. Limits: $x = 0 \to u = 1$, $x = 2 \to u = 5$. So $\tfrac12\int_1^5 u^3du = \tfrac12\left[\tfrac{u^4}{4}\right]_1^5 = \tfrac18(625 - 1) = 78$.
Never returns to $x$.
Route 2. Indefinite: $\tfrac{(x^2+1)^4}{8} + C$. Then $\tfrac{625}{8} - \tfrac18 = 78$. Same answer, same work, different bookkeeping.
Either is right; mixing them is not.
$\displaystyle\int x\sqrt{x-1}\,dx$. Take $u = x - 1$, so $du = dx$ — but the integrand still has a lone $x$.
Step 3 fails, so far.
Rather than abandon it, convert: $x = u + 1$. The integral becomes $\int (u+1)u^{1/2}du = \int (u^{3/2} + u^{1/2})du$.
The stray $x$ is expressed in $u$.
$= \tfrac25 u^{5/2} + \tfrac23 u^{3/2} + C = \tfrac25(x-1)^{5/2} + \tfrac23(x-1)^{3/2} + C$.
Substituted back, as an indefinite integral requires.
$u = x^2 + 1$, $du = 2x\,dx$ — and $2x\,dx$ is the whole numerator.
Step 3 passes.
Limits: $x=0 \to u=1$, $x=1 \to u=2$. The integral is $\int_1^2 \frac{du}{u}$.
$= [\ln|u|]_1^2 = \ln 2 - \ln 1 = \ln 2$.
Match each substitution decision to its justification.
| $u=x^2$ accounts for $du=2x\,dx$ | Adjust by a factor of $1/2$ | Use transformed $u$ bounds and do not return to $x$ | Substitute back to the original variable and add $C$ | |
|---|---|---|---|---|
| $\int 2x\cos(x^2)dx$ | ||||
| $\int x\cos(x^2)dx$ | ||||
| A definite integral converted entirely to $u$ | ||||
| An indefinite antiderivative in $u$ |
Match each substitution situation to the correct setup or conclusion.
| Choose $u=x^2+4$ | $x\,dx=\tfrac16du$ | Use $u$ bounds $1$ and $5$ | Replace the remaining $x$ with $u+1$ | |
|---|---|---|---|---|
| $\int 6x(x^2+4)^5\,dx$ | ||||
| $\int x e^{3x^2}\,dx$ with $u=3x^2$ | ||||
| $\int_0^2 2x(x^2+1)^3\,dx$, $u=x^2+1$ | ||||
| $\int x\sqrt{x-1}\,dx$, $u=x-1$ |
Evaluate $\displaystyle\int_0^1 x\,(x^2)^{4}\,dx$ by substituting $u = x^2$.
Answer:
In $\displaystyle\int_0^{4} 2x\,(x^2+1)^3\,dx$ you substitute $u = x^2 + 1$. What happens to the limits?
| Convert every bound to the new variable | New bound $u=1$ | New bound $u=17$ | Substitute back to $x$ before evaluating | |
|---|---|---|---|---|
| Change from $x$ to $u$ in a definite integral | ||||
| Original lower bound $x=0$ | ||||
| Original upper bound $x=4$ | ||||
| Keep original $x$ bounds |
$F$ is the antiderivative of $(6x + 4)^2$ with $F(0) = \dfrac{64}{18}$. What is $F(2)$?
Answer:
Evaluate $\displaystyle\int_{-3}^{3} x^{3}\sqrt{1 + x^2}\,dx$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $\displaystyle\int x^2\sqrt{x^3 + 5}\,dx$, which substitution works?
You can substitute in definite and indefinite integrals and say when the method cannot apply. Say in your own words why $\int xe^{x^2}dx$ yields and $\int e^{x^2}dx$ does not. That completes Calculus I: the real numbers, limits, the derivative, its applications, and the integral with the theorem that ties the two halves together.
8. Your turn: $\displaystyle\int_0^1 \frac{2x}{x^2+1}\,dx$, step 3