Substitution is the chain rule run backwards, done formally. You replace an awkward part of the integrand with a new variable $u$, so that the whole integral becomes a standard one in $u$.
④ Integrate · ⑤ Substitute back (or use $u$-limits)
Use the substitution $u = x^3 + 2$ to find $\displaystyle\int 6x^2(x^3+2)^4\,dx$.
Use the substitution $u = x + 2$ to find $\displaystyle\int x\sqrt{x+2}\,dx$.
In most exam questions the substitution is given. When it is not, these patterns are the usual choices.
| Integrand contains… | Try | Example |
|---|---|---|
| A bracket or root of a linear expression, times $x$ | $u = $ the linear expression | $\int x\sqrt{x+2}\,dx$, $u = x+2$ |
| A function and (a multiple of) its derivative | $u = $ the function | $\int x e^{x^2}\,dx$, $u = x^2$ |
| $\sqrt{x}$ inside a fraction | $u = \sqrt{x}$ or $u = 1 + \sqrt{x}$ | $\int \dfrac{1}{1+\sqrt{x}}\,dx$ |
| $e^x$ in a fraction | $u = e^x$ or $u = 1 + e^x$ | $\int \dfrac{e^x}{(1+e^x)^2}\,dx$ |
| $\ln x$ together with $\dfrac{1}{x}$ | $u = \ln x$ | $\int \dfrac{(\ln x)^2}{x}\,dx$ |
| $\sqrt{a^2 - x^2}$ | $x = a\sin\theta$ | $\int \sqrt{1-x^2}\,dx$ |
Use $u = 1 + e^x$ to find $\displaystyle\int \frac{e^x}{(1+e^x)^2}\,dx$.
Using $u = \cos x$, find $\displaystyle\int \sin^3 x\,dx$.
Use $u = x + 1$ to evaluate $\displaystyle\int_0^3 \frac{x}{\sqrt{x+1}}\,dx$.
Use $u = 1 + \sqrt{x}$ to show that $\displaystyle\int_0^4 \frac{1}{1+\sqrt{x}}\,dx = 4 - 2\ln 3$.
Evaluate $\displaystyle\int_1^e \frac{(\ln x)^2}{x}\,dx$ using $u = \ln x$.
Sometimes $x$ itself is replaced by a function, such as $x = \sin\theta$. The aim is to use $1 - \sin^2\theta = \cos^2\theta$ to remove a square root.
Use $x = \sin\theta$ to show that $\displaystyle\int_0^1\sqrt{1-x^2}\,dx = \frac{\pi}{4}$.
Replace dx
$dx = \dfrac{du}{du/dx}$, or differentiate $x$ in terms of $u$.
Everything in u
No $x$ may remain before you integrate.
Leftover x
Rearrange the substitution: e.g. $x = u - 2$.
Limits
Convert to $u$-limits; then no need to go back.
Indefinite
Substitute back to $x$ and add $+\,c$.
Function + derivative
Let $u$ be the function.
√(a² − x²)
$x = a\sin\theta$, then $\cos^2\theta$ identity.
Odd power of sin
Save one $\sin x$; let $u = \cos x$.
Use the substitution $u = x^2 - 3$ to find $\displaystyle\int 2x(x^2-3)^6\,dx$.
▶ Show solution
$\dfrac{du}{dx} = 2x$, so $2x\,dx = du$.
$\displaystyle\int u^6\,du = \tfrac{1}{7}u^7 + c = \tfrac{1}{7}(x^2-3)^7 + c$
Use the substitution $u = x - 1$ to find $\displaystyle\int x(x-1)^5\,dx$.
▶ Show solution
$dx = du$ and $x = u + 1$.
$\displaystyle\int (u+1)u^5\,du = \int (u^6 + u^5)\,du = \tfrac{1}{7}u^7 + \tfrac{1}{6}u^6 + c$
$= \tfrac{1}{7}(x-1)^7 + \tfrac{1}{6}(x-1)^6 + c$
Use the substitution $u = x^2$ to find the exact value of $\displaystyle\int_0^1 x e^{x^2}\,dx$.
▶ Show solution
$du = 2x\,dx$, so $x\,dx = \tfrac{1}{2}du$. Limits: $u = 0$ to $u = 1$.
$\displaystyle\int_0^1 \tfrac{1}{2}e^u\,du = \Big[\tfrac{1}{2}e^u\Big]_0^1 = \tfrac{1}{2}(e - 1)$
Use the substitution $u = x - 1$ to evaluate $\displaystyle\int_2^5 \frac{x}{\sqrt{x-1}}\,dx$.
▶ Show solution
$dx = du$, $x = u + 1$. Limits: $x = 2 \Rightarrow u = 1$; $x = 5 \Rightarrow u = 4$.
$\displaystyle\int_1^4 \left(u^{1/2} + u^{-1/2}\right)du = \Big[\tfrac{2}{3}u^{3/2} + 2u^{1/2}\Big]_1^4$
$= \left(\tfrac{16}{3} + 4\right) - \left(\tfrac{2}{3} + 2\right) = \tfrac{14}{3} + 2 = \tfrac{20}{3}$
Use the substitution $u = \sin x$ to find the exact value of $\displaystyle\int_0^{\pi/2}\cos x\,e^{\sin x}\,dx$.
▶ Show solution
$du = \cos x\,dx$. Limits: $x = 0 \Rightarrow u = 0$; $x = \tfrac{\pi}{2} \Rightarrow u = 1$.
$\displaystyle\int_0^1 e^u\,du = e - 1$
Use the substitution $u = \cos x$ to show that $\displaystyle\int_0^{\pi/3}\sin x\cos^2 x\,dx = \frac{7}{24}$.
▶ Show solution
$du = -\sin x\,dx$. Limits: $x = 0 \Rightarrow u = 1$; $x = \tfrac{\pi}{3} \Rightarrow u = \tfrac{1}{2}$.
$\displaystyle\int_1^{1/2} -u^2\,du = \int_{1/2}^1 u^2\,du$ (swapping the limits removes the minus sign).
$= \Big[\tfrac{1}{3}u^3\Big]_{1/2}^1 = \tfrac{1}{3}\left(1 - \tfrac{1}{8}\right) = \tfrac{7}{24}$ ✓
Use the substitution $u = \sqrt{x}$ to show that $\displaystyle\int_1^9 \frac{1}{x + \sqrt{x}}\,dx = 2\ln 2$.
▶ Show solution
$x = u^2$, so $dx = 2u\,du$. Limits: $u = 1$ to $u = 3$.
$\displaystyle\int_1^3 \frac{2u}{u^2 + u}\,du = \int_1^3 \frac{2}{u + 1}\,du$
$= \Big[2\ln(u+1)\Big]_1^3 = 2\ln 4 - 2\ln 2 = 2\ln 2$ ✓
Use the substitution $x = 2\sin\theta$ to find the exact value of $\displaystyle\int_0^{\sqrt{3}}\sqrt{4 - x^2}\,dx$.
▶ Show solution
$dx = 2\cos\theta\,d\theta$. Limits: $x = 0 \Rightarrow \theta = 0$; $x = \sqrt{3} \Rightarrow \sin\theta = \tfrac{\sqrt{3}}{2}$, $\theta = \tfrac{\pi}{3}$.
$\sqrt{4 - 4\sin^2\theta} = 2\cos\theta$, so the integrand becomes $2\cos\theta \times 2\cos\theta = 4\cos^2\theta$.
$\displaystyle\int_0^{\pi/3}4\cos^2\theta\,d\theta = \int_0^{\pi/3}2(1 + \cos 2\theta)\,d\theta = \Big[2\theta + \sin 2\theta\Big]_0^{\pi/3}$
$= \dfrac{2\pi}{3} + \sin\dfrac{2\pi}{3} = \dfrac{2\pi}{3} + \dfrac{\sqrt{3}}{2}$
Use the substitution $u = e^x + 1$ to show that $\displaystyle\int_0^{\ln 3}\frac{e^{2x}}{e^x + 1}\,dx = 2 - \ln 2$.
▶ Show solution
$du = e^x\,dx$, and $e^{2x} = e^x \cdot e^x = (u - 1)e^x$.
Limits: $x = 0 \Rightarrow u = 2$; $x = \ln 3 \Rightarrow u = 4$.
$\displaystyle\int_2^4 \frac{u-1}{u}\,du = \int_2^4\left(1 - \frac{1}{u}\right)du = \Big[u - \ln u\Big]_2^4$
$= (4 - \ln 4) - (2 - \ln 2) = 2 - \ln 2$ ✓
The curve $C$ has equation $y = \dfrac{x^3}{\sqrt{1 + x^2}}$.
(a) Show that the substitution $u = 1 + x^2$ transforms $\displaystyle\int_0^{\sqrt{3}}\frac{x^3}{\sqrt{1+x^2}}\,dx$ into $\displaystyle\frac{1}{2}\int_1^4\frac{u-1}{\sqrt{u}}\,du$. (b) Hence find the exact area of the region bounded by $C$, the $x$-axis and the line $x = \sqrt{3}$. (c) A student claims that $\displaystyle\int_{-\sqrt{3}}^{\sqrt{3}}\frac{x^3}{\sqrt{1+x^2}}\,dx$ gives the total area between $C$ and the $x$-axis for $-\sqrt{3} \leqslant x \leqslant \sqrt{3}$. Explain why the student is wrong and state the correct total area.
▶ Show solution
(a) $du = 2x\,dx$, so $x\,dx = \tfrac{1}{2}du$. Write $x^3\,dx = x^2 \cdot x\,dx = (u - 1)\cdot\tfrac{1}{2}du$.
Limits: $x = 0 \Rightarrow u = 1$; $x = \sqrt{3} \Rightarrow u = 4$. So the integral is $\tfrac{1}{2}\displaystyle\int_1^4 \frac{u-1}{\sqrt{u}}\,du$ ✓
(b) $\tfrac{1}{2}\displaystyle\int_1^4\left(u^{1/2} - u^{-1/2}\right)du = \tfrac{1}{2}\Big[\tfrac{2}{3}u^{3/2} - 2u^{1/2}\Big]_1^4$
$= \tfrac{1}{2}\left[\left(\tfrac{16}{3} - 4\right) - \left(\tfrac{2}{3} - 2\right)\right] = \tfrac{1}{2} \times \tfrac{8}{3} = \tfrac{4}{3}$
The curve is above the axis for $x > 0$, so the area is $\tfrac{4}{3}$.
(c) Replacing $x$ by $-x$ changes the sign of $y$, so the curve has rotational symmetry about the origin. The region for $-\sqrt{3} \leqslant x \leqslant 0$ lies below the axis and its integral is $-\tfrac{4}{3}$; the two parts cancel to give $0$.
The total area is $\tfrac{4}{3} + \tfrac{4}{3} = \tfrac{8}{3}$.