Infi-2 1

Primitive function and indefinite integral

\begin{matrix} Let f be a function \\ Let F be a differentiable function such that F^{'} = f (\frac{d F}{d x} = f) \\ F is then called a primitive function of f \\ Indefinite integral of f is a set of its primitive functions \\ It is denoted as \int f (x) d x = F + C \end{matrix}

\begin{matrix} n \neq 1, \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C \\ \int \frac{1}{x} d x = \ln | x | + C \\ \int a^{x} d x = \frac{a^{x}}{\ln (a)} + c \\ \int e^{x} d x = e^{x} + C \\ \int \cos (x) d x = \sin (x) + C \\ \int \sin (x) d x = - \cos (x) + C \\ \int \frac{1}{1 + x^{2}} d x = \arctan (x) + C \\ \int \frac{1}{\sqrt{1 - x^{2}}} d x = \arcsin (x) + C \\ \int \frac{1}{\cos^{2} (x)} d x = \tan (x) + C \end{matrix}

\begin{matrix} \int (a f + g) (x) d x = a \int f (x) d x + \int g (x) d x \end{matrix}

\begin{matrix} \int \sqrt[6]{x} d x = \int x^{1 / 6} d x = \frac{6}{7} \sqrt[6]{x^{7}} + C \\ \int \frac{7 \cos^{2} (x) + 2 \sin^{2} (x)}{\cos^{2} (x)} d x = \int \frac{5 \cos^{2} (x)}{\cos^{2} (x)} + \frac{2 (\cos^{2} (x) + \sin^{2} (x))}{\cos^{2} (x)} d x = \\ = \int 5 d x + \int \frac{2}{\cos^{2} x} d x = 5 x + 2 \tan (x) + 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 \end{matrix}

\begin{matrix} \int \cos (3 x + 5) d x = \frac{1}{3} \sin (3 x + 5) + C \\ \int e^{3 x} d x = \frac{1}{3} e^{3 x} + C \end{matrix}

\begin{matrix} \sin^{2} (x) + \cos^{2} (x) = 1 \\ \sin (2 x) = 2 \sin (x) \cos (x) \\ \cos (2 x) = 2 \cos^{2} (x) - 1 = 1 - 2 \sin^{2} (x) \\ \sin^{2} (x) = \frac{1 - \cos (2 x)}{2} \\ \cos^{2} (x) = \frac{1 + \cos (2 x)}{2} \\ \sin (x \pm y) = \sin (x) \cos (y) \pm \sin (y) \cos (x) \\ \cos (x \pm y) = \cos (x) \cos (y) \mp \sin (x) \sin (y) \\ \cos (x) \cos (y) = \frac{\cos (x - y) + \cos (x + y)}{2} \\ \sin (x) \sin (y) = \frac{\cos (x - y) - \cos (x + y)}{2} \\ \sin (x) \cos (y) = \frac{\sin (x + y) + \sin (x - y)}{2} \end{matrix}

\begin{matrix} \int \sin (5 x) \cos (2 x) d x = \frac{1}{2} \int (\sin (7 x) + \sin (3 x)) d x = \frac{1}{2} (- \frac{1}{7} \cos (7 x) - \frac{1}{3} \cos (3 x)) + C \\ \int \sin^{4} (x) d x = \int (\sin^{2} (x))^{2} d x = \int {(\frac{1 - \cos (2 x)}{2})}^{2} d x = \\ = \frac{1}{4} \int (1 - 2 \cos (2 x) + \cos^{2} (2 x)) d x = \frac{1}{4} \int (1 - 2 \cos (2 x) + \frac{1}{2} + \frac{\cos (4 x)}{2}) d x = \\ = \frac{1}{4} \int (\frac{3}{2} - 2 \cos (2 x) + \frac{\cos (4 x)}{2}) d x = \frac{1}{4} (\frac{3 x}{2} - \sin (2 x) + \frac{1}{8} \sin (4 x)) + C \end{matrix}

\begin{matrix} \int f (x) g^{'} (x) d x = f (x) g (x) - \int f^{'} (x) g (x) d x \end{matrix}

\begin{matrix} \int x^{4} \ln (x) d x = \int \ln (x) {(\frac{x^{5}}{5})}^{'} d x = \frac{x^{5} \ln (x)}{5} - \int \frac{x^{5}}{5 x} d x = \\ = \frac{x^{5} \ln (x)}{5} + \frac{x^{5}}{25} + C \end{matrix}

\begin{matrix} Order of functions to choose as f, g in integration by parts \\ The higher the place, the better choice it is for f and not g \\ 1. & Logarithmic \\ 2. & Inverse trigonometric \\ 3. & Algebraic \\ 4. & Trigonometric \\ 5. & Exponential \end{matrix}

\begin{matrix} \int \cos (\ln x) d x = x \cos (\ln x) + \int \sin (\ln x) d x \\ \int \sin (\ln x) d x = x \sin (\ln x) - \int \cos (\ln x) d x \\ ⟹ \int \cos (\ln x) d x = x \cos (\ln x) + x \sin (\ln x) - \int \cos (\ln x) d x \\ ⟹ \int \cos (\ln x) d x = \frac{x \cos (\ln x) + x \sin (\ln x)}{2} + C \\ ⟹ \int \sin (\ln x) d x = \frac{x \sin (\ln x) - x \cos (\ln x)}{2} + C \end{matrix}

\begin{matrix} (f (g (x)))^{'} = f^{'} (g (x)) g^{'} (x) \\ ⟹ \int f^{'} (g (x)) g^{'} (x) d x = f (g (x)) + C \\ t = g (x) ⟹ \int f^{'} (t) g^{'} (x) d x = f (t) + C = \int f^{'} (t) d t \\ ⟹ f^{'} (t) g^{'} (x) d x = f^{'} (t) d t ⟹ d t = g^{'} (x) d x \end{matrix}

\begin{matrix} \int \frac{2 x}{1 + x^{2}} d x \\ t = 1 + x^{2} ⟹ d t = (1 + x^{2})^{'} d x = 2 x d x ⟹ d x = \frac{d t}{2 x} \\ ⟹ \int \frac{2 x}{t} \frac{d t}{2 x} = \int \frac{1}{t} d t = \ln | t | + C = \ln | 1 + x^{2} | + C \\ \int \sin^{4} (x) \cos (x) d x \\ t = \sin (x) ⟹ d t = \cos (x) d x \\ ⟹ \int t^{4} d t = \frac{t^{5}}{5} + C = \frac{\sin^{5} (x)}{5} + C \\ \int \arctan (x) d x = x \arctan (x) - \int \frac{x}{1 + x^{2}} d x = \\ = x \arctan (x) - \frac{1}{2} \ln | 1 + x^{2} | + C \end{matrix}

\begin{matrix} \int \sin^{m} (x) \cos^{n} (x) d x \\ Choose function with even power \\ \int \cos^{3} (x) \sin^{2} (x) d x \\ t = \sin (x) ⟹ d t = \cos (x) d x \\ ⟹ \int (1 - t^{2}) t^{2} d t = \int (t^{2} - t^{4}) d t = \frac{t^{3}}{3} - \frac{t^{5}}{5} + C = \frac{\sin^{3} (x)}{3} - \frac{\sin^{5} (x)}{5} + C \\ \int x e^{x^{2}} d x \underset{t = x^{2}}{\underset{⏟}{=}} \frac{1}{2} \int e^{t} d t = \frac{e^{x^{2}}}{2} + C \\ \int x^{3} e^{x^{2}} d x = \frac{x^{2} e^{x^{2}}}{2} - \int x e^{x^{2}} d x = \frac{x^{2} e^{x^{2}}}{2} - \frac{e^{x^{2}}}{2} + C \end{matrix}