Integration Techniques

Integration by Parts

Sometimes you'll face an integral that's a product of two functions, like $x$ and $\cos(x)$ . Simple u-substitution won't work. For these cases, we can use a technique called integration by parts. It comes from reversing the product rule for differentiation.

The product rule states that $(uv)' = u'v + uv'$ . If we integrate both sides, we can rearrange the terms to get the formula for integration by parts.

\int u \, dv = uv - \int v \, du

The key is choosing $u$ and $dv$ correctly. A good rule of thumb is the acronym LIATE, which stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions. You should choose your $u$ as the function that appears first in this list. This choice generally makes the new integral, $\int v \, du$ , simpler than the original one.

Let's try an example: $\int x e^x \, dx$ . Following LIATE, the algebraic function $x$ comes before the exponential function $e^x$ . So, we'll set $u = x$ .

Let	Then
$u = x$	$du = dx$
$dv = e^x \, dx$	$v = \int e^x \, dx = e^x$

Now, we plug these pieces into the formula:

\begin{aligned} \int x e^x \, dx &= (x)(e^x) - \int e^x \, dx \\ &= x e^x - e^x + C \end{aligned}

Trigonometric Integrals

Integrals involving powers of trigonometric functions are common. Solving them usually involves using trigonometric identities to simplify the integrand into a form you can handle with u-substitution.

Let's focus on integrals of the form $\int \sin^m(x) \cos^n(x) \, dx$ . The strategy depends on whether the powers $m$ and $n$ are even or odd.

If the power of cosine is odd, save one cosine factor and use $\cos^2(x) = 1 - \sin^2(x)$ to express the remaining factors in terms of sine. Then, let $u = \sin(x)$ . If the power of sine is odd, the strategy is similar: save one sine factor and use $\sin^2(x) = 1 - \cos^2(x)$ , then let $u = \cos(x)$ .

What if both powers are even? In that case, you'll need to use the half-angle identities to reduce the powers.

\sin^2(x) = \frac{1 - \cos(2x)}{2} \quad \text{and} \quad \cos^2(x) = \frac{1 + \cos(2x)}{2}

Integrals with other trig functions, like tangent and secant, follow similar patterns. You'll often use identities like $\tan^2(x) = \sec^2(x) - 1$ and the fact that the derivative of $\tan(x)$ is $\sec^2(x)$ .

Trigonometric Substitution

When you see integrands with expressions like $\sqrt{a^2 - x^2}$ , $\sqrt{a^2 + x^2}$ , or $\sqrt{x^2 - a^2}$ , a clever technique called trigonometric substitution can be a lifesaver. The idea is to substitute $x$ with a trigonometric function to eliminate the square root, using Pythagorean identities.

The substitution you choose depends on the form of the expression.

Expression	Substitution	Identity Used
$\sqrt{a^2 - x^2}$	$x = a \sin(\theta)$	$1 - \sin^2(\theta) = \cos^2(\theta)$
$\sqrt{a^2 + x^2}$	$x = a \tan(\theta)$	$1 + \tan^2(\theta) = \sec^2(\theta)$
$\sqrt{x^2 - a^2}$	$x = a \sec(\theta)$	$\sec^2(\theta) - 1 = \tan^2(\theta)$

Let's work through an example: $\int \frac{1}{x^2 \sqrt{4 - x^2}} \, dx$ . This integral contains the expression $\sqrt{4 - x^2}$ , which matches the form $\sqrt{a^2 - x^2}$ with $a=2$ . So, we'll use the substitution $x = 2 \sin(\theta)$ . This also means $dx = 2 \cos(\theta) \, d\theta$ .

Substituting these into the integral gives us:

\begin{aligned} &\int \frac{1}{(2\sin\theta)^2 \sqrt{4 - (2\sin\theta)^2}} \cdot 2\cos\theta \, d\theta \\ &= \int \frac{2\cos\theta}{4\sin^2\theta \sqrt{4 - 4\sin^2\theta}} \, d\theta \\ &= \int \frac{2\cos\theta}{4\sin^2\theta \sqrt{4(1 - \sin^2\theta)}} \, d\theta \\ &= \int \frac{2\cos\theta}{4\sin^2\theta \cdot 2\cos\theta} \, d\theta = \int \frac{1}{4\sin^2\theta} \, d\theta \\ &= \frac{1}{4} \int \csc^2(\theta) \, d\theta = -\frac{1}{4} \cot(\theta) + C \end{aligned}

The final step is to convert the answer back to the original variable, $x$ . Since our substitution was $x = 2 \sin(\theta)$ , or $\sin(\theta) = x/2$ , we can draw a right triangle to find the relationship for $\cot(\theta)$ .

From the triangle, we can see that $\cot(\theta) = \frac{\text{adjacent}}{\text{opposite}} = \frac{\sqrt{4 - x^2}}{x}$ .

So, our final answer is:

- \frac{\sqrt{4 - x^2}}{4x} + C

Partial Fraction Decomposition

This technique is used for integrating rational functions, which are fractions where the numerator and denominator are both polynomials. The goal is to break down a complicated rational function into a sum of simpler fractions that are easier to integrate.

For this method to work, the degree of the numerator must be less than the degree of the denominator. If it isn't, you must first perform polynomial long division.

After ensuring the fraction is proper, the process is:

Factor the denominator completely.
For each factor in the denominator, write a term in the partial fraction decomposition. The form of the term depends on the type of factor (linear, repeated linear, quadratic, etc.).
Solve for the unknown coefficients in the numerators of the new fractions.
Integrate the sum of the simpler fractions.

For example, to integrate $\frac{5x-3}{x^2-2x-3}$ , we first factor the denominator: $x^2-2x-3 = (x-3)(x+1)$ . We can then decompose the fraction:

\frac{5x-3}{(x-3)(x+1)} = \frac{A}{x-3} + \frac{B}{x+1}

By multiplying both sides by the original denominator, $(x-3)(x+1)$ , we get $5x-3 = A(x+1) + B(x-3)$ . We can solve for $A$ and $B$ by plugging in strategic values for $x$ . For instance, letting $x=3$ makes the $B$ term zero, giving us $12 = 4A$ , so $A=3$ . Letting $x=-1$ makes the $A$ term zero, giving us $-8 = -4B$ , so $B=2$ .

Our original integral now becomes:

\int \left( \frac{3}{x-3} + \frac{2}{x+1} \right) dx = 3\ln|x-3| + 2\ln|x+1| + C

These techniques expand your integration toolkit significantly. Let's review them.

Ready to test your skills?

Quiz Questions 1/5

When using integration by parts to evaluate the integral $\int x \cos(x) \, dx$ , what is the best choice for $u$ ?

Quiz Questions 2/5

To solve an integral containing the expression $\sqrt{x^2 - a^2}$ , which trigonometric substitution is most appropriate?

Mastering these methods requires practice. Each one is suited for a different kind of problem, and recognizing which technique to apply is a key skill in calculus.

Calculus II