/math/ - \textbf{Primitive function} of function \[f(x)\] over some interval \[x\in[a, b]\] is a function \[F(x)\] whose derivative is the function \[f(x)\] on that interval. ,,\qquad \forall x\in[a, b],\quad F'(x) = f(x) \textbf{Antiderivative} (aka \textbf - Mathematics

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Methods of solving integrals Anonymous June 17, 2022 (Fri) 16:40:29 No. 199

Primitive function of function

f(x)

over some interval

x\in[a, b]

is a function

F(x)

whose derivative is the function

f(x)

on that interval.

\qquad \forall x\in[a, b],\quad F'(x) = f(x)

Antiderivative (aka indefinite integral) of a function

f(x)

is a family of its primitive functions which differ by a constant

C\in\mathbb{R}

:

\qquad \int f(x)\mathrm{d}x = F(x) + C

In a nutshell, integration is the opposite of differentiation:

\qquad \frac{\mathrm{d}\int f(x)\mathrm{d}x}{\mathrm{d}x} = F(x)

\qquad \mathrm{d}\int f(x)\mathrm{d}x = f(x)\mathrm{d}x

\qquad \int\mathrm{d}F(x) = F(x) + C

Solving integrals in general is pretty hard, but there are a lot of established ways to do it. As OP I'll post some of the standard approaches, but this thread is about any kind of integration so feel free to post integrals and theirs solutions.

Most basic methods are:

Using the table of integrals
Using linearity property
Using substitution
Using partial integration
Reducing quadratic to its cannonical form
Partial decomposition

>>

Anonymous June 17, 2022 (Fri) 16:41:17 No. 200

Integration table

Exponents and Powers (special case for

x

is obtained by setting

a=1, b=0

)

$\int (ax + b)\ \mathrm{d}x = a\frac{x^2}{2} + bx + C$	$\int (ax + b)^n\ \mathrm{d}x = \frac{1}{a}\cdot \frac{(ax+b)^{n + 1}}{n + 1} + C\ (n\neq -1)$
$\int\frac{\mathrm{d}x}{ax + b} = \frac{1}{a}\ln\|ax + b\| + C$	$\int e^{ax + b}\mathrm{d}x = \frac{e^b}{a}\cdot e^{ax+b} + C$

Trigonometric:

$\int\sin(\omega x)\mathrm{d}x = -\frac{1}{\omega}\cos(x) + C$	$\int\cos(\omega x)\mathrm{d}x = \frac{1}{\omega}\sin(x) + C$	$\int\tan(\omega x)\mathrm{d}x = \frac{1}{\omega}\ln\left\|\frac{1}{\cos(x)}\right\| + C$
$\int\frac{\mathrm{d}x}{\sin^2(\omega x)} = -\frac{1}{\omega}\cot(\omega x) + C$	$\int\frac{\mathrm{d}x}{\cos^2(\omega x)} = \frac{1}{\omega}\tan(\omega x) + C$

Hyperbolic:

$\int\sinh(x)\mathrm{d}x = \cosh(x) + C$	$\int\cosh(x)\mathrm{d}x = \sinh(x) + C$	$\int\tanh(x)\mathrm{d}x =\ln\cosh(x) + C$
$\int\frac{\mathrm{d}x}{\sinh^2(x)} = -\coth(x) + C$	$\int\frac{\mathrm{d}x}{\cosh^2(x)} = \tanh(x) + C$

Inverse quadratic:

$\int\frac{\mathrm{d}x}{x^2 + a^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$	$\int\frac{\mathrm{d}x}{x^2 - a^2}=\frac{1}{2a}\ln\left\|\frac{x - a}{x + a}\right\| +C$	$\int\frac{\mathrm{d}x}{a^2 - x^2} = \frac{1}{2a}\ln\left\|\frac{a+x}{a-x}\right\| + C$
$\int\frac{\mathrm{d}x}{\sqrt{x^2 + a^2}} = \ln\left\|x + \sqrt{x^2 + a^2}\right\| + C$	$\int\frac{\mathrm{d}x}{\sqrt{x^2 - a^2}} = \ln\left\|x + \sqrt{x^2 - a^2}\right\| + C$	$\int\frac{\mathrm{d}x}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C$

>>

Anonymous June 17, 2022 (Fri) 16:41:58 No. 201

Linearity property

Integration is linear. This means the integral can "go through" a sum or a multiplication with a constant.

\qquad\int\left(\alpha f(x) + \beta g(x)\right)\mathrm{d}x = \alpha\int f(x)\mathrm{d}x + \beta\int g(x)\mathrm{d}x

\qquad\int\left(\sum_{k=1}^n \alpha_k f_k(x)\right)\mathrm{d}x = \sum_{k=1}^n\alpha_k\int f_k(x)\mathrm{d}x

Using this property, it's possible to split an integral into manysmaller integrals which can be solved independently using other methods (e.g. sing table)

Example:

\qquad\int\left(\frac{5}{x} + 7\sin(2x) + x^8\right)\mathrm{d}x = 5\int\frac{\mathrm{d}x}{x} + 7\int\sin(2x)\mathrm{d}x + \int x^8\mathrm{d}x = 5\ln|x| -\frac{7}{2}\cos(2x) + \frac{x^9}{9} +C

>>

Anonymous June 17, 2022 (Fri) 16:42:55 No. 202

Substitution

Let's say we want to integrate

f(x)

in order obtain its primitve function

F(x)

over some interval

[a,b]

:

\qquad \int f(x)\mathrm{d}x = F(x) + C,\quad x\in[a, b]

Now let's suppose we can identify a surjective function

g(x)

that is continuous on

[a,b]

whose derivative on that interval is also continuous.
We can then write.

\qquad \int f(g(x))\cdot g'(x)\mathrm{d}x = \int f(g(x))\ \mathrm{d}g(x) = F(g(x)) + C

In a nushell, we've substituted

x

with

g(x)

and solved the integral in the usual way.
The tricky part is identifying

g'(x)

. Without it, it would not be possible to

Example:
Ok, the based and redpilled way to do this is to notice from the definition of

\mathrm{d}

in the OP is

\mathrm{d}f(x) = f'(x)\mathrm{d}x

.
That means we could write

g'(x)\mathrm{d}x

as

\mathrm{d}g(x)

:

\qquad \int f(g(x))\cdot g'(x)\mathrm{d}x = \int f(g(x))\ \mathrm{d}g(x)

So a funny thing based engineers like to do (which makes mathfags seethe and cope) is this:

\qquad\int\sin(x)\cos(x)\mathrm{d}x = \int\sin(x)\ \mathrm{d}\sin(x) = \frac{\sin^2(x)}{2} + C

Basically, treating

\sin(x)

like

x

and using

\int x\mathrm{d}x = \frac{x^2}{2} + C

.

The cringe and bluepilled way of solving this, of course, is recognizing

\sin(2x) = 2\sin(x)\cos(x)

\qquad\int\sin(x)\cos(x)\mathrm{d}x = \frac{1}{2}\int\sin(2x)\mathrm{d}x = \frac{-\cos(2x)}{4} + C = \frac{2\sin^2(x) - 1}{4} + C = \frac{\sin^2(x)}{2} + \underbrace{\left(-\frac{1}{4} + C\right)}_{\text{constant}} = \frac{\sin^2(x)}{2} + C

>>

Anonymous June 17, 2022 (Fri) 16:43:32 No. 203

Partial integration

This basically allows you to swap what you integrate and what you integrate by.

\qquad\int u\ \mathrm{d}v = uv - \int v\ \mathrm{d}u + C

Usually, you use it to solve an integral with polynomial multiplied by one of the non-polynomial functions.

\qquad \int P_n(x)\cdot \begin{matrix}e^{ax}\\\sin(ax)\\\cos(ax)\\\ln(ax)\\\arcsin(ax)\\\arccos(ax)\end{matrix}\mathrm{d}x

You usually choose

u

to be the non-polynomial because calculating the derivative of it is probably going to be easier than integrating it.
On the other hand, polynomials are easy to both derive and integrate.

Example

You integrate

\int x^2\arccos(x)\mathrm{d}x

by differentiating

\arccos(x)

and by integrating

x^2

:

\begin{aligned} \qquad \int \arccos(x)\cdot x^2 \mathrm{d}x &= \begin{Bmatrix} u = \arccos(x) & \mathrm{d}u = -\frac{\mathrm{d}x}{\sqrt{1- x^2}}\\\mathrm{d}v = x^2\mathrm{d}x & v = \frac{x^3}{3}\end{Bmatrix} = \frac{x^3}{3}\arccos(x)+ \frac{1}{3}\int\frac{x^3}{\sqrt{1-x^2}}\mathrm{d}x +C \\&= \frac{x^3}{3}\arccos(x) - \frac{1}{3}\left(\sqrt{1-x^2} + \frac{\sqrt{(1-x^2)^3}}{3}\right) + C \end{aligned}

As for how you solve

\int\frac{x^3}{\sqrt{1-x^2}}\mathrm{d}x

you do it by substituting

t = \sqrt{1-x^2},\quad x^2 = 1- t^2,\quad \cancel{2}\mathrm{d}x = -\cancel{2}t\mathrm{d}t

:

\qquad \int\frac{x^3}{\sqrt{1-x^2}}\mathrm{d}x = \int\frac{1-t^2}{t}(-t)\mathrm{d}t = t - \frac{t^3}{3} = \boxed{\sqrt{1-x^2} - \frac{\sqrt{(1-x^2)^3}}{3}}

>>

Anonymous June 17, 2022 (Fri) 16:44:12 No. 204

Quadratic trinomial

How do you solve

\int\frac{\mathrm{d}x}{ax^2 + bx + c}

and

\int\frac{\mathrm{d}x}{\sqrt{ax^2 + bx + c}}

? You write

ax^2 + bx + c

in the following way (hint: completing the square by adding and subtracting

\frac{b^2}{4})

:

\begin{aligned} \qquad ax^2 + bx + c &= a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right) = a\left(\left(x^2 + 2\frac{b}{2a}x + \frac{b^2}{4a^2}\right) +\left(- \frac{b^2}{4a^2} + c \right)\right)\\&= a\left(\left(x + \frac{b}{2a}\right)^2 +\left(\sqrt{c - \frac{b}{2a}}\right)^2\right)\\&=a(t^2 + k^2),\quad t= x+ \frac{b}{2a},\quad k=\sqrt{c-\frac{b}{2a}} \end{aligned}

Now the integral reduces to either

\boxed{\frac{1}{a}\int\frac{\mathrm{d}x}{t^2 + k^2} = \frac{1}{ak}\arctan{\frac{t}{k}} + C}

or

\boxed{\frac{1}{a}\int\frac{\mathrm{d}x}{\sqrt{t^2 + k^2}} = \frac{1}{a}\ln\left|t^2 + \sqrt{t^2 + k^2}\right| + C}

>>

Anonymous June 17, 2022 (Fri) 16:44:46 No. 205

Partial fraction decomposition

How do you solve

\int\frac{P(x)}{Q(x)}\mathrm{d}x

where

\ \deg{P(x)} < \deg{Q(x)}

?

First, as a consequence of the fundamental theorem of algebra, any real polynomial can be factored into linear and quadtratic terms.
We will do that with

Q(x)

:

\qquad Q(x) = (x-a_1)^{A_1}(x-a_2)^{A_2}\dots(x - a_m)^{A_m}(x^2 + b_1x + c_1)^{B_1}(x^2 + b_2 x + c_2)^{B_2}\dots(x^2 + b_n x + c_n)^{B_n}

Now employ the partial fraction decomposition:

\qquad\frac{P(x)}{Q(x)} = \sum_{i=1}^m\sum_{j=1}^{A_i}\frac{a_{ij}}{(x-a_i)^j} + \sum_{i=1}^n\sum_{j=1}^{B_i}\frac{b_{ij}x + c_{ij}}{(x^2 + b_ix +)^j}

Then just use the linearity property of the integral.

For example:

\qquad \int\frac{x^2 + 1}{x^5 - 3x^4 + x^3 + 7x^2 - 6x - 8}\mathrm{d}x = \int\frac{x^2 + 1}{(x-2)(x +1)^2(x^2-3x + 4)}\mathrm{d}x = \int \frac{a_{11}}{x-2}\mathrm{d}x + \int\frac{a_{21}}{x+1}\mathrm{d}x + \int\frac{a_{22}}{(x+1)^2}\mathrm{d}x + \int\frac{b_{11} x + c_{11}}{x^2 - 3x + 4} \mathrm{d}x

The constants

a_{ij},b_{ij}, c_{ij}\in\mathbb{R}

have to be found e.g. using the Heaviside cover up method.

>>

Anonymous August 10, 2023 (Thu) 15:09:44 No. 464

cool nad thanks

$\int\frac{\mathrm{d}x}{x^2 + a^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$	$\int\frac{\mathrm{d}x}{x^2 - a^2}=\frac{1}{2a}\ln\left\|\frac{x - a}{x + a}\right\| +C$	$\int\frac{\mathrm{d}x}{a^2 - x^2} = \frac{1}{2a}\ln\left\|\frac{a+x}{a-x}\right\| + C$
$\int\frac{\mathrm{d}x}{\sqrt{x^2 + a^2}} = \ln\left\|x + \sqrt{x^2 + a^2}\right\| + C$	$\int\frac{\mathrm{d}x}{\sqrt{x^2 - a^2}} = \ln\left\|x + \sqrt{x^2 - a^2}\right\| + C$	$\int\frac{\mathrm{d}x}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C$