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    Copyright (C) 2013 Frédéric Wang

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Basic Integration

Ideas

If F is a primitive of f then \int_{a}^b f(x) d x = F(b) - F(a), provided the integrand and integral have no singularities on the path of integration. As a consequence, we can use a table of derivatives:

Examples

Using linearity and elementary power rule:

Using linearity and sine/cosine:

Using Leibniz rule, Chain rule and Exponential/Power:

Using the inverse function rule and tangent:

Using linearity, reciprocal rule and logarithm:

Using the power rules and chain rule:

Integration by Substitution

Idea

Do a substitution x = g(y) to simplify an integral. We have \frac{dx}{dy} = g'(y) and \int f(x) dx = \int f(g(y)) g'(y) dy: this is the chain rule!

Example 1

Let y = x^2 - 1, dy = 2 x dx then we get

We find the same result as

Example 2

Consider

Let y = x^3 - 2, dy = 3x^2 dx so

Example 3

Consider

Let u = -\frac{1}{v^2}, du = \frac{2}{v^3} dv:

Example 4

Consider

Let t = 2 \sin \theta and so dt = 2 \cos \theta d \theta. We have

We have -1 < \cos \theta < 0 on \left[\frac{3\pi}{4},\frac{7\pi}{6}\right] and so

Example 5

Consider

We first write

Now doing the substitution y = \frac{x - \frac{3}{2}}{\sqrt{\frac{5}{4}}}, dy = \frac{dx}{\sqrt{\frac{5}{4}}} we get

But we have already met this integral in the basic methods:

Example 6

Consider

We do the substitution t = \tan\left(\frac{\theta}{2}\right). We have d \theta = \frac{2 dt}{1+t^2}, \cos \theta = \frac{1-t^2}{1+t^2} and \sin \theta = \frac{2t}{1+t^2}. Also, we have

and so

and finally

Integration by parts

Idea

Given a two function f, g we have

this is Leibniz rule!

Example 1

Consider

We consider f(x) = x^2 + x + 1 and g(x) = \frac{e^{-3x}}{-3} that is f'(x) = 2x+1 and g'(x) = e^{-3x}. The integral can be written \int f g' dx and hence we get

We now consider h(x) = 2x+1, h'(x)=2 the second integral can be written \int h g' dx and hence

We now recognize g' in the last integral and so

Finally,

Example 2

We let f(x) = 2\ln x + 7\arctan x and g(x) = \frac{x^3}{3}. Hence f'(x) = \frac{2}{x} + \frac{7}{1+x^2} and g'(x) = x^3. The integration by parts gives:

The integral can be written

Finally

Example 3

Consider

We do an integration by parts with f(\theta) = \sin(3 \theta) and g'(\theta) = e^{2 \theta}:

We repeat a similar integration by parts on the remaining integral:

We note that we found the initial integral I. We get:

Finally,

Integration by Partial Fraction Decomposition

Idea

Given a rational function \frac{P}{Q}, write its partial fraction decomposition and integrate each term.

Example 1

Clearly, we have 6x^5-18x^4+18x^3-6x^2 = 6 x^2 (x^3-3x^2+3x-1) and 1 is a trivial root of the second factor so we can factor it by (x-1) and obtain:

Hence we search a decomposition of

for some constants A, B, C, D, E \in \mathbb{R}. We have

Finally,

Example 2

The denominator can be written u^5-4u^4+14u^3-20u^2+25u = u(u^4-4u^3+14u^2-20u+25). Evaluating the second factor at 0, \pm1, \pm2 we don't find any trivial roots. Hence we try to convert it to a depressed quartic by setting u = v - \frac{-4}{4} = v + 1. We find (u^4-4u^3+14u^2-20u+25) = v^4 + 8v^2 + 16 = (v^2 + 4)^2 = (u^2-2u+5)^2. The discriminant of u^2-2u+5 is 4(1-5) < 0 so we have the decomposition in irreducible factors:

We now try to find A, B, C, D, E \in \mathbb{R} such that

From the two last equalities, we immediately get D = 2 and E = -2. Finally,

We have

hence

Example 3

By polynomial long division we find that

and so

The integral of the first term is:

First we search trivial roots for the denominator. If we evaluate t^4+2t^3-t-2 at t=0,-1,1,-2,2 we find respectively -2, -2, 0, 0, 28. So we can factor X^4+2X^4-X-2 by (X - 1)(X + 2). We get

The polynomial X^2+X+1 is irreducible on \mathbb{R} because its discrimant is 1 - 4 = -3 < 0. Hence we search a decomposition of

for some constants A, B, C, D \in \mathbb{R}. We have

So

The integral of the two first terms is easy to compute:

The last term can be split into two parts:

We have

and

Finally we get

Alternatively, we can write

where t^\pm = \frac{-1\pm i \sqrt{3}}{2} are the complex roots of t^2+t+1. We have

We have

After simplification, we get

as we already found above.

Contour Integration

Idea

Consider contour integral \oint_\gamma f(z) d z along paths in complex plane to deduce integral along the real line.

Example 1

Consider the integral

Write z = e^{i \theta}, d z = i z d \theta. The integral becomes

where \gamma is the unit circle traversed counterclockwise. f : z \mapsto \frac{3(z^4+1)}{2 z^2 (z - 2 i)\left(z - \frac{i}{2}\right)} has two singularities inside that circle: 0 and \frac{i}{2}. Hence the integral is the 2 \pi i times the sum of the residues of f at these points:

Hence

Example 2

We want to know the value of the integral

For that purpose, we introduce the function

We note that x \mapsto \Im(f(x)) is the integrand above. We also note that the poles of f are z^\pm = -1 \pm 2 i. By choosing the right contour, one can show that

We have

Hence

Finally