Integration by Substitution

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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, $du=2xdx$.

$\begin{aligned} \int_{-1}^{0}\frac{2x}{\left(2+x^{2}\right)^{3}}dx& =\int_{-1}^{0}2x(2+x^{2})^{-3}dx=\int_{3}^{2}u^{-3}du  \\ &=-\frac{1}{2}u^{-2}\bigg|_{3}^{2}=-\frac{1}{2}\bigg[\frac{1}{4}-\frac{1}{9}\bigg]=-\frac{5}{72} \end{aligned}$