Strategies of Integration

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Substitution

The most basic trick for doing integrals is $u$-substitution. The simplest uses involve taking $u$ to be the inside part of a composition of functions, and have $du$ or a multiple of $du$ appearing in the integral as well. More sophisticated uses involve clever choices of $u$. There is no single rule for how to pick $u$. It takes practice to get a feel for it. With definite integrals, the limits of integration must be changed, or the antiderivative must be computed separately.

Standard example:

$\displaystyle\int x \sin\left(x^2+3\right) \,dx \overset{\fbox{$\,\,\,\, u\,=\,x^2+3\\\,\,\, du\,=\,2x\,dx\\ \frac{1}{2}du\,=\,x\,dx$}\\}{=} \frac{1}{2} \int\sin u\,du = -\frac{1}{2}\cos u + C = -\frac{1}{2}\cos\left(x^2+3\right) + C$

Example of substitution on a definite integral, with two ways to solve.

$\displaystyle\int_1^2\frac{dx}{(3x-5)^2}\,dx \overset{\fbox{$\,\,\,\, u\,=\,3x-5\\\,\,\, du\,=\,3x\,dx\\ \frac{1}{3}du\,=\,x\,dx\\u(1)\,=\,3\cdot 1-5\,=\,-2\\u(2)\,=\,3\cdot 2-5\,=\,1$}\\}{=}\frac{1}{3}\int_{-2}^1 u^{-2}\,du =-\frac{1}{3}u^{-1}\left|\begin{array}{c} ^{1} \\ _{-2} \end{array}\right .=-\frac{1}{3}\left(\frac{1}{1}-\frac{1}{-2}\right)=-\frac{1}{3}\frac{3}{2}=-\frac{1}{2}$

$\hspace{1in}$ -----------------${\bf OR}$-------------------

$\displaystyle\int_1^2\frac{dx}{(3x-5)^2}\,dx$
$\qquad\left(\text{ First find the antiderivative: }\displaystyle\int\frac{dx}{(3x-5)^2}\,dx \overset{\fbox{$\,\,\,\, u\,=\,3x-5\\ \,\,\, du\,=\,3x\,dx\\ \frac{1}{3}du\,=\,x\,dx$}\\}{=}\frac{1}{3}\int u^{-2}\,du=-\frac{1}{3}u^{-1}+C=-\frac{1}{3(3x-5)}+C\right)$

$\displaystyle\int_1^2\frac{dx}{(3x-5)^2}\,dx=-\frac{1}{3(3x-5)}\left|\begin{array}{c} ^{2} \\ _{1} \end{array}\right .=-\frac{1}{3}\left(\frac{1}{3\cdot 2-5}-\frac{1}{3\cdot{1}-5}\right)-\frac{1}{3}\left(\frac{1}{1}-\frac{1}{-2}\right)=-\frac{1}{3}\frac{3}{2}=-\frac{1}{2}$

Two examples with clever choices of $u$:

$\displaystyle\int e^{\sqrt x}\,dx\overset{\fbox{$ u\,=\,\sqrt x \\du\,=\,\frac{1}{2\sqrt x}\,dx\\2\sqrt x\,du\,=\,dx\\2u\,du\,=\,dx$}\\}{=}\int 2e^u u\,du \qquad\text{ and now use integration by parts, and rewrite in terms of } x$
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$\displaystyle\int x\sqrt{x-1}\,dx \overset{\fbox{$\,\,\,\, u\,=\,x-1\\\,\,\, du\,=\,dx\\ x\,=\,(u+1)$}\\}{=} \int(u+1)\sqrt u\,du=\int(u^{3/2}+u^{1/2})\,du\qquad\text{ which you can integrate, then rewrite in terms of } x $

The Fundamental Theorem of Calculus

The Indefinite Integral and the Net Change

Substitution

Area Between Curves

Volumes

Integration by Parts

Integrals of Trig Functions

Trig Substitutions

Partial Fractions

Strategies of Integration

Improper Integrals

Differential Equations

Models of Growth

Infinite Sequences

Infinite Series

Integral Test

Comparison Tests

Convergence of Series with Negative Terms

The Ratio and Root Tests

Strategies for testing Series

Power Series

Representing Functions as Power Series

Taylor and Maclaurin Series

Applications of Taylor Polynomials

Partial Derivatives

Multiple Integrals

Iterated Integrals over Rectangles

Double Integrals over General Regions

Substitution

Standard example:

Example of substitution on a definite integral, with two ways to solve.

Two examples with clever choices of $u$: