Calculus review: integration II

3) Trig substitutions
Often an integrand can be much simplified by a trigonmetric identity, such as $$\sin ^2 +\cos^2=1,$$ or $$1+\tan^2=\sec^2.$$  For instance, consider the integral

$$\int_a^b \frac{dx}{\sqrt{1-x^2}}.$$

We know that $$1-\sin^2=\cos^2,$$ so what if we let $$x=\sin \theta?$$

We then have the squareroot as cos, and $$dx=\cos \theta d\theta,$$ so our integral is now

$$\int_{\arccos a}^{\arccos b}\frac{\cos\theta d\theta}{\cos\theta}.$$

Simplifying, we get
$$\int_{\arccos a}^{\arccos b}d\theta=\theta \bigg |_{\arccos a}^{\arccos b}.$$

Done.

Let's try another one.  What if we want to do

$$\int _a^b \frac{dx}{1+x^2}?$$

Let's try letting $$x=\tan\theta,$$ which means $$dx=\sec^2\theta d\theta.$$  We then have

$$\int _{\arctan a}^{\arctan b} \frac{\sec^2\theta d\theta }{\sec^2 \theta}.$$  Simplifying, we get

$$\int _{\arctan a}^{\arctan b}d\theta=\theta\bigg| _{\arctan a}^{\arctan b}.$$

Done.  Notice that, as with integrating by substitution, you have to remember to calculate dx and also change the limits of integration.

4) Partial fractions
Suppose we want to integrate what is called a rational function: a ratio of two polynomials.  For instance,
$$\int \frac{x^3+2x^2-x}{x^2-1}dx.$$

We might first think of making a substitution either with $$u=x^2-1$$ or perhaps a trig substitution, but we end up being better served if we notice that the integrand can be written as

$$\frac{x^2}{x-1}+\frac{x}{x+1}.$$  We can then integrate the first term of this sum using the substitution $$u=x-1,\;du=dx:$$
$$\int\frac{x^2}{x-1}dx=\int \frac{(u+1)^2}{u}du=\int \frac{u^2+2u+1}{u}du.$$

Dividing all of the terms by u gives us
$$\int \big[u+2+\frac{1}{u}\big]du,$$

which can easily be integrated.  The second term of the sum can be integrated with the substitution $$u=x+1,\;du=dx.$$

What we have done here is decomposed one fraction into a sum of fractions: basically, reversed what you do when you add fractions and find the least common denominator.  This technique is called partial fractions.  Usually you can just reconstruct the form you need with a bit of guess and check, by writing the numerators of the terms in the decomposition as a+bx and c+dx, for instance, and solving for a, b, c, and d by adding the terms and demanding equality of the coefficients of each power of x with the original fraction.