LocusBlog

U substitution, and the limits people forget to change

U substitution undoes the chain rule. The chain rule says $\frac{d}{dx}F(g(x)) = F'(g(x))g'(x)$, so reading that equation from right to left,

$$\int f(g(x))\,g'(x)\,dx = \int f(u)\,du = F(u) + C, \qquad u = g(x)$$

where $F$ is any antiderivative of $f$. The bookkeeping is the substitution $u = g(x)$ together with $du = g'(x)\,dx$, and the whole method is the claim that those two replacements turn an integral you cannot do into one you can.

The condition is that the integrand really does contain the derivative of the inside function as a factor, up to a constant. A constant is fine, because constants pull out of integrals. An extra $x$ is not fine, and an integrand that does not have the inner derivative in it at all is not a substitution problem.

Choosing u

Look for a composition, and take $u$ to be the inside. Then check whether $g'(x)$ is present as a factor. If it is off by a constant multiple, fix the constant and continue.

$$\int x\sqrt{x^2+9}\;dx$$

The composition is $\sqrt{\phantom{x}}$ applied to $x^2+9$, so $u = x^2+9$ and $du = 2x\,dx$. The integrand has $x\,dx$, not $2x\,dx$, which is a factor of two away, so $x\,dx = \tfrac12\,du$:

$$\int x\sqrt{x^2+9}\;dx = \frac{1}{2}\int \sqrt{u}\;du = \frac{1}{2}\cdot\frac{2}{3}u^{3/2} + C = \frac{1}{3}(x^2+9)^{3/2} + C$$

Every indefinite integral can be checked by differentiating the answer, and there is no reason not to. The chain rule on $\tfrac13(x^2+9)^{3/2}$ gives $\tfrac13 \cdot \tfrac32 (x^2+9)^{1/2}\cdot 2x = x\sqrt{x^2+9}$, the original integrand.

The definite case, done both ways

$$\int_0^4 x\sqrt{x^2+9}\;dx$$

There are two legitimate finishes, and they are not equally safe.

Convert back. Do the indefinite integral as above, restore the $x$, and evaluate at the original limits:

$$\left[\frac{1}{3}(x^2+9)^{3/2}\right]_0^4 = \frac{1}{3}\left(25^{3/2} - 9^{3/2}\right) = \frac{125 - 27}{3} = \frac{98}{3}$$

Change the limits. $u = x^2+9$ sends $x = 0$ to $u = 9$ and $x = 4$ to $u = 25$, and the integral becomes an integral in $u$ that never mentions $x$ again:

$$\frac{1}{2}\int_9^{25}\sqrt{u}\;du = \frac{1}{2}\cdot\frac{2}{3}\left[u^{3/2}\right]_9^{25} = \frac{125 - 27}{3} = \frac{98}{3}$$

Both give $98/3 = 32.667$. A midpoint Riemann sum with two million subintervals gives $32.6666667$, which is the same number.

Prefer changing the limits. The converted-back version asks you to carry an expression in $x$ through the algebra and then remember it was in $x$; the changed-limits version finishes inside the new variable and is done. The advantage grows with the ugliness of the substitution - for a trigonometric substitution, undoing $u$ means drawing a triangle, and having already turned the limits into numbers saves all of it. The one time to convert back is when you were asked for an antiderivative anyway.

Changing the variable but not the limits

This is the error the second method invites, and it is worth seeing the size of it. Substitute $u = x^2+9$, produce $\tfrac12\int\sqrt u\,du$ correctly, and then carry the old limits $0$ and $4$ into the new integral:

$$\frac{1}{2}\int_0^4 \sqrt{u}\;du = \frac{1}{3}\left[u^{3/2}\right]_0^4 = \frac{8}{3} = 2.667$$

The right answer is $98/3 = 32.667$. The wrong one is twelve times too small, and it is wrong in a way that arithmetic alone will never catch, because $8/3$ is a perfectly reasonable-looking number.

Estimate the integral before you trust it. On $[0,4]$ the integrand $x\sqrt{x^2+9}$ rises from $0$ to $4\sqrt{25} = 20$, so the area under it over an interval of width $4$ has an average height somewhere between those. The answer $98/3$ has average height $8.17$. The answer $8/3$ has average height $0.67$, for a curve that ends at $20$. A single glance at the size of the integrand rejects it.

The habit that prevents the error is to write the new limits at the moment you write $du$, in the same breath, rather than at the end. An integral sign with $u$ underneath it and $x$-limits on it is a statement that is simply false, and if you never write it down you can never evaluate it.

When the substitution is not sitting in plain sight

Sometimes the inner derivative is only there after you rewrite. Nothing about $\int \tan x\,dx$ suggests a substitution until you write $\tan x$ as $\sin x/\cos x$; then $u = \cos x$ gives $du = -\sin x\,dx$ and the integral is $-\int du/u = -\ln|\cos x| + C$.

Sometimes there is more of the old variable than $du$ can absorb, and the fix is to solve the substitution equation backwards.

$$\int x^5\sqrt{x^3+1}\;dx$$

Take $u = x^3+1$, so $du = 3x^2dx$. The integrand has $x^5 = x^3 \cdot x^2$, so $x^2dx$ is accounted for by $\tfrac13 du$, and the leftover $x^3$ is handled by the substitution itself: $x^3 = u - 1$. Now nothing in $x$ remains.

$$\frac{1}{3}\int (u-1)\sqrt{u}\;du = \frac{1}{3}\int \left(u^{3/2} - u^{1/2}\right)du = \frac{2}{15}u^{5/2} - \frac{2}{9}u^{3/2} + C$$

So the answer is $\frac{2}{15}(x^3+1)^{5/2} - \frac{2}{9}(x^3+1)^{3/2} + C$. Differentiate it to check: the two terms give $\frac13 u^{3/2}\cdot 3x^2 - \frac13 u^{1/2}\cdot 3x^2 = x^2\sqrt u\,(u-1)$, and $u - 1 = x^3$, so that is $x^5\sqrt{x^3+1}$. Numerically, at $x = 1.3$ the derivative of the antiderivative is $6.638777$ and the integrand is $6.638777$.

Substitution is the first integration technique because almost everything later is built on top of it - integration by parts usually ends in a substitution, partial fractions produce a pile of them, and trigonometric substitution is a substitution with a triangle attached. Run indefinite substitution problems until choosing $u$ is instant, then switch to definite substitution where the limits are the thing being tested. The study guide section works both finishes side by side, and if the reverse direction feels shaky, the chain rule is the forward version of the same statement.

References