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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