Substitution Rule
If \(u=g(x)\) is a differentiable function whose range is an interval \(I\) and if \(f\) is continuous on \(I\), then \(\int f(g(x))g'(x)dx=\int f(u)du.\)
Note that the assumptions guarantee that the integrands on both sides of this equality are continuous functions.
TIP: Substitute terms that are:
- raised to a high power: \(\int(3x^{2}+2x)(x^{3}+x^{2})^{3}dx\) let \(u=x^{3}+x^{2}\)
- under a root: \(\int4x^{3}\sqrt[5]{x^{4}-1}dx\) let \(u=x^{4}-1\)
- in the power of the exponential \(\int e^{x^3+x}(3x^2+1)dx\) let \(u=x^{3}+x\)
- within the logarithmic function: \(\int-3(4x^{3}-x^{2})\mathrm{ln}(-3x^{4}+x^{3})dx\) let \(t=-3x^{4}+x^{3}\)
Example: Integrate \(\int3(8z+1)e^{4z^{2}+z}dz.\)
Solution: Let \(u=4z^2 + z\). Then, \(du = (8z+1)dz\).
\(\begin{aligned}&\int3(8z+1)e^{4z^2+z}dz=\int3e^udu=3\int e^udu\\&=3e^u+C=3e^{4z^2+z}+C\end{aligned}\)
Example: Integrate \(\int\frac x{\sqrt{1-4x^{2}}}dx.\)
Solution: Let \(u=1-4x^2\). Then, \(du=-8xdx\), but rewrite it as \(-\frac{1}{8}du=xdx\).
\(\begin{aligned}
\int{\frac{x}{\sqrt{1-4x^{2}}}}dx& =\int x(1-4x^{2})^{-\frac{1}{2}}dx=-\frac{1}{8}\int u^{-\frac{1}{2}}du \\
&=-\frac{1}{4}u^{\frac{1}{2}}+C=-\frac{1}{4}(1-4x^{2})^{\frac{1}{2}}+C
\end{aligned}\)
Definite Integral by Substitution
If \(g'(x)\) is continuous on \([a,b]\) and if \(f\) is continuous on the range of \(u=g(x)\), then
$$\begin{aligned}
\int_{a}^{b}f(g(x))g'(x)dx& =\int_{g(a)}^{g(b)}f(u)du \\
&=F(u)\Big|_{g(a)}^{g(b)}=F(g(b))-F(g(a))
\end{aligned}$$
where \(F\) is an antiderivative of \(f\). (Note that an alternative solution is provided to the example below for illustration of this formula.)
Computing a Definite Integral by Substitution
Step 1: Solve the integral as an indefinite integral.
Step 2: Use the result of the indefinite integral, and evaluate it over the interval of integration.
Alternatively, change the limits of integration when converting the integral to the new variable u (see example below).
Example: Integrate \(\int_{-1}^0\frac{2x}{\left(2+x^2\right)^3}dx.\)
Solution: Step 1
Let \(u=2+x^2\). Then \(du=2xdx\).
\(\begin{aligned}
\int{\frac{2x}{(2+x^{2})^{3}}}dx& =\int2x(2+x^{2})^{-3}dx=\int u^{-3}du \\
&=-\frac{1}{2}u^{-2}+C=-\frac{1}{2}(2+x^{2})^{-2}+C
\end{aligned}\)
Step 2
\(\begin{aligned}
\int_{-1}^{0}\frac{2x}{\left(2+x^{2}\right)^{3}}dx& =-\frac{1}{2}(2+x^{2})^{-2}\Bigg|_{-1}^{0} \\
&=\biggl[-\frac{1}{2}\bigl(2+(0)^{2}\bigr)^{-2}\biggr]-\biggl[-\frac{1}{2}\bigl(2+(-1)^{2}\bigr)^{-2}\biggr] \\
&=\frac{1}{2}\Big[-(2)^{-2}+(3)^{-2}\Big]=\frac{1}{2}\Big[-\frac{1}{4}+\frac{1}{9}\Big]=-\frac{5}{72}
\end{aligned}\)
Alternative solution
Let \(u=2+x^2\). Then, \(du=2xdx\).
\(\begin{aligned}
\int_{-1}^{0}\frac{2x}{\left(2+x^{2}\right)^{3}}dx& =\int_{-1}^{0}2x(2+x^{2})^{-3}dx=\int_{3}^{2}u^{-3}du \\
&=-\frac{1}{2}u^{-2}\bigg|_{3}^{2}=-\frac{1}{2}\bigg[\frac{1}{4}-\frac{1}{9}\bigg]=-\frac{5}{72}
\end{aligned}\)