Substitution: the chain rule backwards
Tonight's idea
The chain rule says \(\frac{d}{dx}F(g(x)) = F'(g(x))\,g'(x)\). Read it backwards:
\(\displaystyle\int F'(g(x))\,g'(x)\,dx = F(g(x)) + C.\)
So an integrand that is "a function of something, times that something's derivative" is a chain rule in disguise. The bookkeeping: name the inner function \(u = g(x)\). Then \(\frac{du}{dx} = g'(x)\), which we write as \(du = g'(x)\,dx\). (Not a fraction being split; a reminder that a small step \(dx\) moves \(u\) by \(g'(x)\,dx\).) Replace \(g(x)\) by \(u\) and \(g'(x)\,dx\) by \(du\), and the \(x\) is gone.
Try \(\int 2x\cos(x^2)\,dx\). The inner function is \(x^2\); its derivative \(2x\) is sitting right there. Put \(u = x^2\), \(du = 2x\,dx\):
\[\int 2x\cos(x^2)\,dx = \int \cos u\,du = \sin u + C = \sin(x^2) + C.\]
Check by differentiating: \(\cos(x^2) \cdot 2x\). Correct. For a definite integral, move the limits too: if \(x\) runs from \(a\) to \(b\), then \(u\) runs from \(g(a)\) to \(g(b)\), and you never return to \(x\). Why is the area unchanged? The \(u\)-axis is a stretched copy of the \(x\)-axis. A strip of width \(dx\) becomes a strip of width \(du = g'(x)\,dx\), wider by the factor \(g'(x)\), and its height drops from \(F'(g(x))\,g'(x)\) to \(F'(u)\), shorter by the same factor. Strip by strip, the area is the same.
Watch one: \(\int_0^1 \frac{x}{1 + x^2}\,dx\)
Inner function \(1 + x^2\), derivative \(2x\). We only have \(x\), so fix the constant: \(du = 2x\,dx\) means \(x\,dx = \tfrac12\,du\). Limits: \(x = 0\) gives \(u = 1\), \(x = 1\) gives \(u = 2\).
\[\int_0^1 \frac{x}{1 + x^2}\,dx = \frac12\int_1^2 \frac{du}{u} = \frac12\Big[\ln u\Big]_1^2 = \frac12\ln 2 \approx 0.347.\]
Two habits. A missing constant can be supplied, but a missing \(x\) cannot: \(\int \cos(x^2)\,dx\) does not yield to this method. And once the limits are in \(u\), the answer is a number; nothing is substituted back.
See an area under \(1/x\) built from rectangles (there from 1 to \(e\)) at mathdetours.com/demos/riemann.html.
Your turn
- Find (a) \(\int (3x + 2)^4\,dx\) and (b) \(\int x\,e^{x^2}\,dx\). Name the substitution each time, and check by differentiating.
- Evaluate \(\int_0^{\pi/2} \sin^3 x\,\cos x\,dx\), moving the limits to \(u\).
- (a) Write \(\tan x\) as \(\sin x / \cos x\) and find \(\int \tan x\,dx\). (b) Evaluate \(\int_0^1 x\sqrt{1 - x}\,dx\) with \(u = 1 - x\). Here the leftover \(x\) must be rewritten as \(1 - u\).
- (a) With \(u = a - x\), show that \(\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx\) for any \(f\). (b) Hence show \(\int_0^{\pi/2} \sin^2 x\,dx = \int_0^{\pi/2} \cos^2 x\,dx\), add the two, and find both.
Stretch: substituting the other way
Evaluate \(\displaystyle\int_0^1 \sqrt{1 - x^2}\,dx\) by putting \(x = \sin\theta\): the new variable is introduced, not spotted. Say what \(dx\) becomes, what the limits become, and check the answer against a picture of the integrand.
Answers
- (a) \(u = 3x + 2\), \(du = 3\,dx\): \(\tfrac13\int u^4\,du = \tfrac{1}{15}(3x+2)^5 + C\). (b) \(u = x^2\), \(du = 2x\,dx\): \(\tfrac12\int e^u\,du = \tfrac12 e^{x^2} + C\). Differentiating each gives back the integrand.
- \(u = \sin x\), \(du = \cos x\,dx\); \(u\) runs from 0 to 1. \(\int_0^1 u^3\,du = \tfrac14\).
- (a) \(u = \cos x\), \(du = -\sin x\,dx\): \(\int \tan x\,dx = -\int \frac{du}{u} = -\ln|\cos x| + C\). (b) \(x = 1 - u\), \(dx = -du\), limits 1 to 0; flipping them absorbs the sign: \(\int_0^1 (1-u)\sqrt u\,du = \int_0^1 (u^{1/2} - u^{3/2})\,du = \tfrac23 - \tfrac25 = \tfrac{4}{15}\).
- (a) \(u = a - x\), \(dx = -du\), limits \(a \to 0\): \(\int_0^a f(a - x)\,dx = -\int_a^0 f(u)\,du = \int_0^a f(u)\,du\), which is \(\int_0^a f(x)\,dx\) with the variable renamed. (b) \(\sin(\tfrac\pi2 - x) = \cos x\), so the two integrals are equal; call each \(I\). Their sum is \(\int_0^{\pi/2}(\sin^2 x + \cos^2 x)\,dx = \pi/2\), so \(I = \pi/4\).
Stretch. \(x = \sin\theta\), \(dx = \cos\theta\,d\theta\); \(x = 0 \to \theta = 0\), \(x = 1 \to \theta = \pi/2\). Then \(\sqrt{1 - \sin^2\theta} = \cos\theta\) (positive on this range), so the integral is \(\int_0^{\pi/2}\cos^2\theta\,d\theta = \pi/4\) by exercise 4. Picture: \(y = \sqrt{1 - x^2}\) on \([0, 1]\) is a quarter of the unit circle, area \(\pi/4\). Same number, two routes.