Infi-2 2-3

Variable substitution in integrals #lemma

\begin{matrix} When there is a pattern in the integral, being function f (x) \\ and it’s derivative \\ We will use variable substitution: \\ Let t = f (x) \\ t^{'} d t = f^{'} (x) d x ⟹ d t = f^{'} (x) d x \\ For example: \\ \int \frac{7 \arctan (x)}{x^{2} + 1} d x \\ (\arctan (x))^{'} = \frac{1}{x^{2} + 1} \\ ⟹ \int \frac{7 \arctan (x)}{x^{2} + 1} d x \underset{f (x) = \arctan (x)}{\underset{⏟}{=}} 7 \int f (x) f^{'} (x) d x = \\ \underset{t = f (x)}{\underset{⏟}{=}} 7 \int t d t = \frac{7 t^{2}}{2} + C = \frac{7 \arctan^{2} (x)}{2} + C \\ Another example: \\ \int 6 \ln (\sin x) \cos^{3} (x) d x \\ Let t = \sin x ⟹ d t = \cos (x) d x \\ ⟹ \int 6 \ln (\sin x) \cos^{3} (x) d x = 6 \int \ln (t) (1 - t^{2}) d t \\ Let f (t) = \ln (t) ⟹ f^{'} (t) = \frac{1}{t} \\ Let g^{'} (t) = 1 - t^{2} ⟹ g (t) = t - \frac{t^{3}}{3} \\ ⟹ 6 \int \ln (t) (1 - t^{2}) d t = 6 (\ln (t) (t - \frac{t^{3}}{3})) - 6 \int (1 - \frac{t^{2}}{3}) d t = \\ = 6 t \ln (t) - 2 t^{3} \ln (t) - 6 t + \frac{2 t^{3}}{3} + C = \\ = 6 \sin (x) \ln (\sin x) - 2 \sin^{3} (x) \ln (\sin x) - 6 \sin (x) + \frac{2 \sin^{3} (x)}{3} + C \end{matrix}

\begin{matrix} We can also substitute in a more complex way: \\ g (t) = f (x) ⟹ g^{'} (t) d t = f^{'} (x) d x \\ For example: \\ \int \sqrt{7 - x^{2}} d x \\ \int \sqrt{7 - x^{2}} d x = \sqrt{7} \int \sqrt{1 - {(\frac{x}{\sqrt{7}})}^{2}} d x \\ Let \sin t = \frac{x}{\sqrt{7}} ⟹ \cos (t) d t = \frac{d x}{\sqrt{7}} ⟹ d x = \sqrt{7} \cos (t) d t \\ ⟹ \sqrt{7} \int \sqrt{1 - {(\frac{x}{\sqrt{7}})}^{2}} d x = \sqrt{7} \int \sqrt{1 - \sin^{2} (x)} d x = \sqrt{7} \int \sqrt{\cos^{2} (x)} d x = \\ = 7 \int \cos^{2} (t) d t = 7 \int \frac{1 + \cos (2 t)}{2} d t = \frac{7}{2} (t + \frac{\sin (2 t)}{2}) + C = \\ = \frac{7}{2} (\arcsin (\frac{x}{\sqrt{7}}) + \frac{\sin (2 \arcsin (\frac{x}{\sqrt{7}}))}{2}) + C \end{matrix}

\begin{matrix} \int f (x) d x = F (x) + C ⟹ \int f (a x + b) d x = \frac{1}{a} F (a x + b) + C \\ Proof: \\ Let t = a x + b ⟹ d t = a d x \\ ⟹ \int f (a x + b) d x = \int \frac{f (t)}{a} d t = \frac{1}{a} F (t) + C = \frac{1}{a} F (a x + b) + C \\ Example: \\ \int 3^{x} d x = \frac{3^{x}}{\ln (3)} + C \\ ⟹ \int 3^{18 x - 5} d x = \frac{\frac{1}{18} 3^{18 x - 5}}{\ln (3)} + C \end{matrix}

\begin{matrix} \int \frac{p (x)}{q (x)} d x \\ There are usually three steps: \\ 1. Division of polynomials \\ 2. Partial fraction decomposition \\ 3. Integration of partial fractions \end{matrix}

\begin{matrix} Division of polynomials: \\ 1.1 Divide highest degree of nominator by highest degree of denominator \\ 1.2 Multiply result by denominator \\ 1.3 Subtract result from nminator \\ 1.4 Repear until degree of nominator is greater or equal to degree of denominator \\ For example: \\ \frac{x^{9} + 6 x^{2} + 6}{x^{4} - 1} \\ \frac{x^{9}}{x^{4}} = x^{5} \cdot (x^{4} - 1) ⟹ x^{9} - x^{5} \\ x^{9} + 6 x^{2} + 6 - x^{9} - x^{5} = x^{5} + 6 x^{2} + 6 \\ \frac{x^{5}}{x^{4}} = x \cdot (x^{4} - 1) ⟹ x^{5} - x \\ x^{5} + 6 x^{2} + 6 - x^{5} - x = 6 x^{2} + x + 6 \\ ⟹ x^{9} + 6 x^{2} + 6 = x^{5} + x + \frac{6 x^{2} + x + 6}{x^{4} - 1} \end{matrix}

\begin{matrix} Partial fraction decomposition: \\ Any polynomial can be decomposed into product of non-decomposable \\ polynomials of smaller degree \\ Non-decomposable polynomial is a polynomial that cannot be represented as a product \\ of polynomials of smaller degrees \\ A non-trivial example would be: \\ x^{4} + 1 = x^{4} + 2 x^{2} + 1 - 2 x^{2} = (x^{2} + 1)^{2} - 2 x^{2} = (x^{2} + 1 - \sqrt{2} x) (x^{2} + 1 + \sqrt{2} x) \\ In our case: \\ x^{4} - 1 = (x^{2} - 1) (x^{2} + 1) = (x - 1) (x + 1) (x^{2} + 1) \\ ⟹ \frac{6 x^{2} + x + 6}{x^{4} - 1} = \frac{6 x^{2} + x + 6}{(x - 1) (x + 1) (x^{2} + 1)} = \frac{.}{x - 1} + \frac{.}{x + 1} + \frac{.}{x^{2} + 1} \\ How do we calculate nominators now? \\ If denominator is linear, nominator will be a constant \\ If denominator is of second degree, nominator will be linear \\ ⟹ \frac{6 x^{2} + x + 6}{x^{4} - 1} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C x + D}{x^{2} + 1} \\ 6 x^{2} + x + 6 = A (x + 1) (x^{2} + 1) + B (x - 1) (x^{2} + 1) + (C x + D) (x^{2} - 1) \\ Let x = 1 \\ 13 = 4 A ⟹ A = \frac{13}{4} \\ Let x = - 1 \\ 11 = - 4 B ⟹ B = - \frac{11}{4} \\ A + B + C = 0 ⟹ C = - \frac{2}{4} \end{matrix}

\begin{matrix} \frac{p (x)}{(x - 1)^{3} (x + 1)} = \frac{A x^{2} + B x + C}{(x - 1)^{3}} + \frac{D}{x + 1} \\ \frac{A x^{2} - 2 A x + A}{(x - 1)^{3}} = \frac{A (x - 1)^{2}}{(x - 1)^{3}} = \frac{A}{x - 1} \\ ⟹ \frac{A x^{2} + B x + C}{(x - 1)^{3}} = \frac{A}{x - 1} + \frac{B x + 2 A x + C - A}{(x - 1)^{3}} \\ \frac{B x + 2 A x - B - 2 A}{(x - 1)^{3}} = \frac{B + 2 A}{(x - 1)^{2}} \\ ⟹ \frac{B x + 2 A x + C - A}{(x - 1)^{3}} = \frac{B + 2 A}{(x - 1)^{2}} + \frac{- B - C - A}{(x - 1)^{3}} \\ ⟹ \frac{A x^{2} + B x + C}{(x - 1)^{3}} = \frac{A}{x - 1} + \frac{B + 2 A}{(x - 1)^{2}} + \frac{- (A + B + C)}{(x - 1)^{3}} = \\ = \frac{A_{1}}{x - 1} + \frac{A_{2}}{(x - 1)^{2}} + \frac{A_{3}}{(x - 1)^{3}} \end{matrix}

\begin{matrix} This results in a following table of denominators and their decompositions \\ \begin{array}{cc} \frac{p (x)}{a x + b} & \frac{A}{a x + b} \\ \frac{p (x)}{(a x + b)^{k}} & \sum_{i = 1}^{k} \frac{A_{i}}{(a x + b)^{i}} \\ \frac{p (x)}{a x^{2} + b x + c} & \frac{A x + B}{a x^{2} + b x + c} \\ \frac{p (x)}{(a x^{2} + b x + c)^{k}} & \sum_{i = 1}^{k} \frac{A_{i} x + B_{i}}{(a x^{2} + b x + c)^{i}} \end{array} \end{matrix}

\begin{matrix} Integration of partial fractions: \\ \int \frac{A}{a x + b} d x = \frac{A}{a} \ln | a x + b | + C \\ \int \frac{A}{(a x + b)^{k}} d x = \frac{A (a x + b)^{1 - k}}{1 - k} + C \\ \int \frac{A x + B}{a x^{2} + b x + c} d x = Generally not in this course, but it is written below \\ \int \frac{A x + b}{(a x^{2} + b x + c)^{k}} d x = Generally not in this course \\ Specific cases of \frac{A x + B}{a x^{2} + b x + c} : \\ \int \frac{1}{(x + a)^{2} + b^{2}} d x = \frac{1}{b} \int \frac{1}{{(\frac{x + a}{b})}^{2} + 1} = \frac{1}{b} \arctan (\frac{x + a}{b}) + C \\ \int \frac{p^{'} (x)}{p (x)} d x = \int \frac{1}{p} d p = \ln | p (x) | + C \end{matrix}

\begin{matrix} \int \frac{3 x + 1}{x^{2} + 2 x + 6} d x = \frac{3}{2} \int \frac{(2 x + \frac{2}{3})}{x^{2} + 2 x + 6} d x = \\ = \frac{3}{2} \int \frac{2 x + 2}{2 x^{2} + 2 x + 6} d x - 2 \int \frac{1}{x^{2} + 2 x + 6} d x = \\ = \frac{3}{2} \ln | 2 x^{2} + 2 x + 6 | - 2 \int \frac{1}{x^{2} + 2 x + 6} d x \\ x^{2} + 2 x + 5 = (x + 1)^{2} + 5 \\ ⟹ \int \frac{1}{x^{2} + 2 x + 6} d x = \int \frac{1}{(x + 1)^{2} + (\sqrt{5})^{2}} d x = \frac{1}{\sqrt{5}} \arctan (\frac{x + 1}{\sqrt{5}}) + C \end{matrix}

\begin{matrix} (a x^{2} + b x + c)^{'} = 2 a x + b \\ \frac{A x + B}{a x^{2} + b x + c} = \frac{A}{2 a} \cdot \frac{2 a x + b + \frac{2 a B}{A} - b}{a x^{2} + b x + c} = \frac{A}{2 a} \frac{2 a x + b}{a x^{2} + b x + c} + (B - \frac{A b}{2 a}) \frac{1}{a x^{2} + b x + c} \\ \frac{1}{a x^{2} + b x + c} = \frac{1}{{(\sqrt{a} x + \frac{b}{2 \sqrt{a}})}^{2} + (c - \frac{b^{2}}{4 a})} = \dots \\ \int \frac{A x + B}{a x^{2} + b x + c} d x = \frac{A}{2 a} \ln | a x^{2} + b x + c | + \frac{(A b - 2 B a) \arctan (\frac{2 a x + b}{\sqrt{4 a c - b^{2}}})}{a \sqrt{4 a c - b^{2}}} + C \end{matrix}