Strategies of Integration

By using suitable substitutions, we can convert integrals involving quadratic expressions, or square roots of quadratic expressions, into trig integrals that can be solved using the methods in this learning module.

For integrals involving $a^2 + x^2$ , use the substitution $x=a\tan(\theta),\qquad dx = a \sec^2(\theta)\,d\theta$ and the identity $a^2+x^2 = a^2 \sec^2(\theta)$ .

For integrals involving $a^2-x^2$ , use the substitution $x=a\sin(\theta),\qquad dx=a\cos(\theta)\, d\theta$ and the identity $a^2-x^2 = a^2 \cos^2(\theta)$ .

For integrals involving $x^2-a^2$ , use the substitution $x=a\sec(\theta),\qquad dx = a\sec(\theta)\tan(\theta)\,d\theta$ and the identity $x^2-a^2 = a^2 \tan^2(\theta)$ .

No matter which substitution we use, we end up with an integral involving $\theta$ and trig functions of $\theta$ . We can use triangles to evaluate those trig functions.

For integrals involving a quadratic expression that is not a sum of squares, "complete the square": $x^2 + 2bx + c = (x+b)^2 + c-b^2$ and then use the substitution $u=x+b$ .

Example: Evaluate

$\int {\sqrt{x^2+6x+10}}\, dx.$

Solution: We first complete the square to get

$x^2+6x+10=(x+3)^2+1$ and then use the

$u$ -substitution

$u\,=\,x+3, \qquad du\,=\,dx.$ This gives us

$\int {\sqrt{x^2+6x+10}}\, dx= \int{\sqrt{(x+3)^2+1}}\, dx= \int \sqrt{u^2+1}\,du.$ The substitution

$u\,=\,\tan(\theta),\qquad du\,=\,\sec^2(\theta)\,d\theta$ yields

$\int \sqrt{u^2+1}\,du= \int \left(\sqrt{\tan^2(\theta)+1}\right)\sec^2(\theta)\,d\theta =\int \sec^3(\theta)\,d\theta.$ From Example 1,

$\int \sec^3(\theta)\,d\theta= \frac{\sec(\theta)\tan(\theta) + \ln\lvert\sec(\theta)+\tan(\theta)\rvert}{2}+C,$ so what we're left to do is to substitute back

$u$ and then

$x$ . Notice that

$\sec(\theta)=\sqrt{1+u^2}$ , by using right triangles. All together, we have

$\hspace{-3cm} \int {\sqrt{x^2+6x+10}}\, dx =\frac{\left(\sqrt{1+u^2}\right)u+ \ln\Bigl\lvert\left(\sqrt{1+u^2}\right)+u\Bigr\rvert}{2}+C$

$\qquad\qquad =\frac{\left(\sqrt{1+(x+3)^2}\right)(x+3)+ \ln\Bigl\lvert\left(\sqrt{1+(x+3)^2}\right)+x+3\Bigr\rvert}{2}+C$

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

Trig Substitution