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Computing Integrals by Substitution

Many integrals are most easily computed by means of a change of variables, commonly called a u-substitution.

Example

Let's compute $\displaystyle\int\! 2x(x^2-1)^4\, dx$ by making the substitution

\begin{eqnarray*} u&=&x^2-1\\ du&=&2x\, dx \end{eqnarray*}

Then
\begin{displaymath} \int 2x(x^2-1)^4\, dx=\int (x^2-1)^4(2x\, dx)=\int u^4\, du=\frac{u^5}{5}+C=\frac{(x^2-1)^5}{5}+C.\end{displaymath}

We may check this result by differentiating using the Chain Rule:
\begin{displaymath}\frac{d}{dx}\left(\frac{(x^2-1)^5}{5}+C\right)=\frac{5(x^2-1)^4}{5}(2x) =2x(x^2-1)^4.\qquad\qquad \surd\end{displaymath}

The substitution method amounts to applying the Chain Rule in reverse:

To compute $\displaystyle\int\! f(g(x))g'(x)\, dx$, we let

\begin{eqnarray*} u&=&g(x)\\ du&=&g'(x)\, dx \end{eqnarray*}

Then
\begin{displaymath}\int f(g(x))g'(x)\, dx=\int f(u)\, du=F(u)=F(g(x))\end{displaymath}

where F is an antiderivative of f.

Example

To compute $\displaystyle\int\! \sin (2x)\cos (2x)\, dx$, let

\begin{eqnarray*} u&=&\sin (2x)\\ du&=&2\cos (2x)\, dx \end{eqnarray*}

Then
\begin{displaymath}\int \sin (2x)\cos (2x)\, dx=\int\frac{1}{2}\sin (2x)[2\cos (... ...nt \frac{1}{2}u\, du=\frac{1}{4}u^2+C=\frac{1}{4}\sin^2 (2x)+C.\end{displaymath}

With practice, you will often be able to write down the result immediately.

Example

We can evaluate $\displaystyle\int\! \frac{dx}{(4x-3)^2}$ by letting

\begin{eqnarray*} u&=&4x-3\\ du&=&4\, dx\quad\longrightarrow\quad dx=\frac{1}{4}\, du. \end{eqnarray*}

Then

\begin{displaymath}\int \frac{dx}{(4x-3)^2}=\int \frac{\frac{1}{4}\, du}{u^2}=-\frac{1}{4u}+C=\frac{-1}{4(4x-3)}+C.\end{displaymath}

It is not always apparent until you try it whether or not a substitution will work.

Example

To compute $\displaystyle\int\! x\sqrt{x-3}\, dx$, we will try

\begin{eqnarray*} u&=&x-3\quad\longrightarrow\quad x=u+3\\ du&=&dx \end{eqnarray*}

So
\begin{eqnarray*} \int x\sqrt{x-3}\, dx&=&\int (u+3)\sqrt{u}\, du=\int \left(u^{... ...ac{2}{5}u^{5/2}+2u^{3/2}+C=\frac{2}{5}(x-3)^{5/2}+2(x-3)^{3/2}+C \end{eqnarray*}

We can also compute a definite integral using a substitution.

Example

Let's evaluate $\displaystyle\int^2_0\! xe^{x^2}\, dx$. Let

\begin{eqnarray*} u&=&x^2\\ du&=&2x\, dx \end{eqnarray*}

First, we will compute the indefinite integral:
\begin{displaymath}\int xe^{x^2}\, dx=\int \left(\frac{1}{2}e^{x^2}\right)(2x\, dx)=\int\frac{1}{2}e^u\, du=\frac{1}{2}e^u+C=\frac{1}{2}e^{x^2}+C.\end{displaymath}

Now we have two approaches for the definite integral:

Approach 1                       Approach 2
Substitute back to                       Change the limits of integration:
the original variable:                       Since $u=x^2$,
$\int xe^{x^2}\, dx=\frac{1}{2}e^u +C$                       $u=0$ when $x=0$
$\qquad\qquad~=\frac{1}{2}e^{x^2}+C$.                       and $u=4$ when $x=2$.
So $\int xe^{x^2}\, dx=\frac{1}{2}e^{x^2}\vert^2_0$      $\Longrightarrow$      $\frac{1}{2}(e^4-1)$      $\Longleftarrow$      $\int^2_0 xe^{x^2}\, dx=\int^4_0 \frac{1}{2}e^u\, du=\frac{1}{2}e^u\vert^4_0$

Thus, we find that

\begin{displaymath}\int^2_0 xe^{x^2}\, dx=\frac{1}{2}(e^4-1).\end{displaymath}

Approach 2 works provided certain conditions on f and g are met:

\begin{displaymath}\int^b_a f(g(x))\, dx=\int^{g(b)}_{g(a)} f(u)\, du\end{displaymath}

if
  1. g' is continuous on [a, b].
  2. f is continuous on the set of values taken by g on [a, b].

Substitutions are useful or necessary for a huge range of integrals. You will find yourself either implicitly or explicitly using a substitution in virtually every integral you compute!

Key Concepts

The substitution method amounts to applying the Chain Rule in reverse:

To compute $\displaystyle\int\! f(g(x))g'(x)\, dx$, we let $u = g(x)$ and $du = g'(x) dx$

Then $\int f(g(x))g'(x)\, dx = \int f(u)\, du = F(u) = F(g(x))$, where F is an antiderivative of f.

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