Integration

Suppose that $F(x)$ and $G(x)$ are antiderivatives of $f(x)$ and $g(x)$, respectively

	Formula
Constant Multiplication	$\int cf(x)dx=c\int f(x)dx$	$c\in \mathbb{R}$
Sum / Difference	$\int \left(f(x)\pm g(x)\right)dx=\int f(x)dx\pm \int g(x)dx$
(2.4) Linearity	$\int (\alpha f(x)+\beta g(x))dx=\alpha \int f(x)dx+\beta \int g(x)dx$
Power Rule	$\int x^{r}dx=\frac{x^{r+1}}{r+1}+C$	$-1\ne r\in \mathbb{R}$
	$\int \frac{1}{x}dx=\ln |x|+C$
Exponential	$\int a^{x}dx=\frac{a^x}{\ln a}$ (Special Case $\int e^{x}dx=e^x$)	$a\ne 1,a>0$
U-Substitution	$\int f(g(x))g^{\prime }(x)=F(g(x))+C$

• ln
  • $\int \ln xdx=x\ln x-x+C$
• Trigonometric
  • $\int \sin xdx=-\cos x+C$
  • $\int \cos xdx=\sin x+C$
  • $\int \tan xdx=\ln \left|\cos x\right|+C$
  • $\int \cot xdx=\ln \left|\sin x\right|+C$
• Reciprocal Trigonometric
  • $\int \frac{1}{\sin x}dx=\ln \left|\tan \frac{x}{2}\right|+C$
  • $\int \frac{1}{\cos x}dx=-\ln \left|\tan \left(\frac{x}{2}-\frac{\pi }{4}\right)\right|+C$
  • $\int \frac{1}{{\sin }^{2}x}dx=-\cot x$
  • $\int \frac{1}{{\cos }^{2}x}dx=\tan x$
• Inverse Trigonometric
  • $\int \arcsin xdx=x\arcsin x+\sqrt{1-x^2}+C$
  • $\int \arctan xdx=x\arctan x-\frac{1}{2}\ln (x^2+1)+C$
• Der. of Inverse Trigonometric
  • $\int \frac{1}{1+x^2}dx=\arctan x+C$
  • $\int \frac{1}{\sqrt{1-x^2}}dx=\arcsin x+C$
  • $\int \frac{1}{\sqrt{a^2+x^2}}dx=\arcsin \frac{x}{a}+C$

some exmaple

• (2.8a) $\int x\ln xdx=\frac{x^2\ln x}{2}-\frac{x^2}{4}+C$

• (2.9) $\int \ln xdx=x\ln x-x+C$

• (2.10) $\int {\ln }^{2}xdx=x{\ln }^{2}x-2x\ln x+2x+C$

• (e2.12) $I_m=\int \frac{1}{(x^2+a^2{)}^m}dx=\{\begin{array}{l}{I}_1=\frac{1}{a}\arctan \frac{x}{a}+C\\ I_{m+1}=\frac{1}{2m{a}^2}\cdot \frac{x}{(x^2+a^2{)}^m}+\frac{2m-1}{2m{a}^2}{I}_m\end{array}$

• (by power rule)

  • $\int \frac{1}{\sqrt{x}}dx=2\sqrt{x}+C$
  • $\int 0dx=C$
  • $\int 1dx=\int dx=x+C$

• $\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C$ (for $\alpha \ne -1$)todo

• $\int \frac{c}{ax+b}dx=\frac{c}{a}\ln \left|ax+b\right|+C$

Read More https://en.wikipedia.org/wiki/Integration_by_reduction_formulae https://en.wikipedia.org/wiki/Lists_of_integrals

Integration by Parts

(2.6)

$$\int f(x)g^{\prime }(x)dx=f(x)g(x)-\int f^{\prime }(x)g(x)dx$$

By substitution,

\begin{align} u=f(x)&\implies du=f'(x)\,dx \\ dv=g'(x)\,dx &\implies v=\int g'(x) \, dx=g(x) \end{align} $$ we get:

\int u,dv =uv-\int v,du

> The **LIATE** rule: choose $u=f(x)$ as the function that appears first in LIATE and choose $dv=g'(x)\,dx$ as the last one. (Logarithmic, Inverse-trig, Algebric, Trig, Exponential) # Substitution - Assumptions: - $f$ is continuous on $I$ - $g$ is continuously differentiable on $J$ - $g(J)\subseteq I$ (i.e. image of $g$ is subset of $I$, so that $f \circ g$ is defined) - Substitute & Evaluate: $$\displaystyle \int f(g(x))g'(x)\, dx={\color{gray}\left[\begin{align} u &= g(x) \\ du &= g'(x)dx \end{align}\right]}=\int f(u) \, du$$ - Return to $x$ by substitute $u=g(x)$ ##### Logarithmic Integration - $\displaystyle\int \frac{f'(x)}{f(x)} \, dx=\ln|f(x)|+C$ (Logarithmic Integration, a substitution of $u=f(x)$ p136) ### Version 2 - Substitute & Evaluate:

\int f(x) , dx={\color{gray}\left[\begin{align} x &= \varphi(t) \ dx &= \varphi’(t)dt \end{align}\right]}=\int f(\varphi(t))\varphi’(t) , dt

- return to $x$ by substitute $t=\varphi^{-1}(x)$ $\int f(x) \, dx=\int f(g(t))g'(t) \, dt\Big\vert_{t=g'(x)}$ - (2.5) $f(\alpha x+\beta)$ form (where $\alpha\neq 0$) - $\displaystyle\int f(x) \, dx=F(x)+C\implies\displaystyle\int f(\alpha x+\beta) \, dx=\frac{1}{\alpha}F(\alpha x+\beta)+C$ #### Trigonometric substitution | Integrand containing | Substitution | Identity | | Result of Substitution | After Simplification | | -------------------- | -------------------------------------------------------------------------- | ------------------- | ----------------------------------------------------- | ---------------------- | -------------------- | | ${a^2-x^2}$ | $x=a\sin t$ <br>$dx=a\cos t \,dt$ <br>$t=\arcsin {\frac {x}{a}}$ | $\sin^2t+\cos^2t=1$ | $-\frac{\pi}{2}\leq t\leq \frac{\pi}{2}$ | $a^2(1-\sin^2 t)$ | $a^2\cos^2t$ | | ${a^2+x^2}$ | $x=a\tan t$ <br>$dx=a\sec^2 t \,dt$ <br>$t=\arctan {\frac {x}{a}}$ | $\tan^2t+1=\sec^2t$ | $-\frac{\pi}{2}<t<\frac{\pi}{2}$ | $a^2(1+\tan^2t)$ | $a^2\sec ^2t$ | | ${x^2-a^2}$ | $x=a\sec t$<br>$dx=a\sec t\tan t \,dt$<br>$t=\text{arcsec} {\frac {x}{a}}$ | $\sec^2t-1=\tan^2t$ | $0\leq t<\frac{\pi}{2}$ or $\pi\leq t<\frac{3\pi}{2}$ | $a^2(\sec^2t-1)$ | $a^2\tan ^2t$ | | Integrands | Substitutions | | ---------------------------------------------- | ------------------------------------------------------------------------- | | $a^{2}+(f(x))^{2}$ | $f(x)=a\tan\theta$ or<br>$a\cot\theta$ or<br>$a\sinh\theta$ | | $a^{2}-(f(x))^{2}$ | $f(x)=a\sin\theta$ or <br>$a\cos\theta$ | | $(f(x))^{2}-a^{2}$ | $f(x)=a\sec\theta$ or <br>$a \csc$ or <br>$a\cosh\theta$ | | $a+f(x)$ | $f(x)=a\tan^{2}\theta$ or <br>$a\cot^{2}\theta$ or <br>$a\sinh^{2}\theta$ | | $a-f(x)$ | $f(x)=a\sin^{2}\theta$ or <br>$a\cos^{2}\theta$ | | $f(x)-a$ | $f(x)=a\sec^{2}\theta$ or <br>$a \csc^2 \theta$ or <br>$\cosh^{2}\theta$ | | $ax-x^{2}$ | $x=a\sin^{2}\theta$ or<br>$a \cos² 0$ | | $(a-x)(x-b)$ | $x=a\cos^{2}\theta+b\sin^{2}\theta$ | | $\displaystyle\frac{a^{n}-x^{n}}{a^{n}+x^{n}}$ | $x^{n}=a^{n}\cos\theta$ | #### Tangent half-angle substitution - $\displaystyle \sin x={\frac {2t}{1+t^{2}}}$ - $\displaystyle\cos x={\frac {1-t^{2}}{1+t^{2}}}$ - $\displaystyle dx={\frac {2}{1+t^{2}}} dt$ - $\int f(\cos x)\sin x \, dx$ then we can substitute $t=\cos x$ - $\int f(\sin x)\cos x \, dx$ then we can substitute $t=\sin x$ - $\displaystyle\int \sin^{m}x\cos^{n}x \, dx$ ($m,n$ are nonnegative integers) - $m$ is odd: - $m=2k+1$ - $\sin ^m x=\sin^{2k+1}x=(\sin^2x)^{k}\sin x=(1-\cos^{2} x)^{k}\sin x$ - $n$ is odd: - $n=2k+1$ - $\cos ^n x=\cos^{2k+1}x=(\cos^2x)^{k}\cos x=(1-\sin^{2} x)^{k}\cos x$ - $m,n$ are even: - #todo # Rational Functions - $\displaystyle \frac{a}{(x-p)^k}$ (where $a,p \in \mathbb{R},\,k \in \mathbb{N}$) - $x-p=t$ and $dx=dt$ - $\displaystyle \int \frac{a}{(x-p)^{k}} \, dx=\begin{cases}{ a\ln|t|+C} & k=1 \\ a \frac{t^{1-k}}{1-k}+C & k\geq 2 \end{cases}=\begin{cases}{ a\ln|x-p|+C} & k=1 \\ a \frac{(x-p)^{1-k}}{1-k}+C & k\geq 2 \end{cases}$ - $\displaystyle \frac{ax+b}{(x^2+px+q)^k}$ (where $a,b,p,q \in \mathbb{R},\,k \in \mathbb{N}$ and $x^2+px+q$ has no real roots) - Step 1: - $\displaystyle ax+b=\frac{a}{2}(2x+p)+b-\frac{ap}{2}$ (note $(2x+p)=(x^2+px+q)'$) - $\int \frac{ax+b}{(x^2+px+q)^k} \, dx= \frac{a}{2}\int \frac{2x+p}{(x^2+px+q)} \, dx+\left( b-\frac{ap}{2} \right)\int \frac{dx}{(x^2+px+q)^k}$ - Step 2: (evaluate the first integral) - $u=x^2+px+q$ and $du=(2x+p)dx$ - $\displaystyle\int \frac{2x+p}{(x^2+px+q)^{k}} \, dx=\int \frac{du}{u^k}=\begin{cases}{ \ln|u|+C} & k=1 \\ \frac{u^{1-k}}{1-k}+C & k\geq 2 \end{cases}$ - $\displaystyle\int \frac{2x+p}{(x^2+px+q)}dx=\ln(x^2+px+q)+C$ - $\displaystyle\int \frac{2x+p}{(x^2+px+q)^k}dx=\frac{1}{1-k}(x^2+px+q)^{1-k}+C$ (where $k\geq 2$) - Step 3: (evaluate the second integral) - $\displaystyle \int \frac{dx}{(x^2+px+q)^k}$ - $(x^2+px+q)=\left( x+\frac{p}{2} \right)^2+q-\frac{p^2}{4}$ ([[Completing the square]]) - $(x^2+px+q)=\left( x+\frac{p}{2} \right)^2+d^2$ - substitute: $x+\frac{p}{2}=t$ and $dx=dt$ - $\displaystyle \int \frac{dx}{(x^2+px+q)^k}=\int \frac{dx}{\left(\left( x+\frac{p}{2} \right)^2+d^2\right)^k}=\int \frac{dt}{(t^2+d^2)^k}$ - $\displaystyle I_{k}=\int \frac{dt}{(t^2+d^2)^k}$ (see [[#examples|Examples]] e2.12) - $\displaystyle I_{1}=\frac{1}{d}\arctan \frac{t}{d}+C$ - $\displaystyle I_{k+1}=\frac{1}{2kd^2}\cdot \frac{t}{(t^2+d^2)^k}+\frac{2k-1}{2kd^2}I_k$ ### Partial Fraction Decomposition Decomposition of a [[Rational function]] $N(x)/D(x)$ into partial fractions 1. Divide when improper: When $N(x)/D(x)$ is [[Rational function|improper]] (i.e. $\deg N \geq \deg D$), [[Polynomial#polynomial-long-division|divide]] the denominator into the numerator to obtain $\displaystyle\frac{N(x)}{D(x)}=Q(x)+\frac{N_{1}(x)}{D(x)}$ where $\deg N_{1}(x)< \deg D(x)$. Then apply Steps 2, 3, and 4 to the proper rational expression $N_{1}(x)/D(x)$ 2. Factor denominator: Completely factor the denominator into factors of the form $(px+q)^m$ and $(ax^2+bx+c)^n$ where $ax^2+bx+c$ is irreducible 3. Linear factors: For each factor of the form $(px + q)^m$, the partial fraction decomposition must include the following sum of $m$ fractions $\displaystyle\frac{A_{1}}{(px+q)}+\frac{A_{2}}{(px+q)^2}+\dots+\frac{A_{m}}{(px+q)^m}$ 4. Quadratic factors: For each factor of the form $(ax^2 + bx + c)^n$, the partial fraction decomposition must include the following sum of $n$ fractions. $\displaystyle \frac{B_{1}x+C_{1}}{ax^2+bx+c}+\frac{B_{2}x+C_{2}}{(ax^2+bx+c)^2}+\dots+\frac{B_{n}x+C_{n}}{(ax^2+bx+c)^n}$ # Definite Integrals | | Formula | | | -------------- | --------------------------------------------------------------------------------- | ---------------------------------------- | | Additivity | $\displaystyle\int_a^b f(x) \; dx = \int_a^c f(x) \; dx \, + \int_c^b f(x) \; dx$ | $a<c<b$ and $f$ is integrable on $[a,b]$ | | Shift Property | $\displaystyle\int_a^b f(x) \; dx = \int_{a-c}^{b-c} f(x+c) \; dx$ | | | | $\displaystyle\int_0^a f(x) \; dx = \int_{0}^{a} f(a-x) \; dx$ | | | | $\displaystyle\int ^b_{a}\alpha \, dx=\alpha(b-a)$ | | | | $\displaystyle\int ^a_{a}f(x) \, dx=0$ | | | | | | - Cavalieri's Quadrature Formula $\displaystyle\int ^b_{a} x^n\, dx=\begin{cases}\displaystyle{\frac{b^{n+1}-a^{n+1}}{n+1}}&{n=1}\\\displaystyle{\ln\left(\frac{b}{a}\right)}&{n=-1}\end{cases}$ (where $a,b,x\in\mathbb{R}$) ## Integration by Parts (2.10) Integration by Parts for Definite Integrals

\displaystyle\int_{a}^{b}f(x)g’(x),dx={\Big [}f(x)g(x){\Big ]}_{a}^{b}-\int _{a}^{b}f’(x)g(x),dx

By substitution, $$\small\displaystyle{\begin{align} u=f(x)&\implies du=f'(x)\,dx \\ dv=g'(x)\,dx &\implies v=\int g'(x) \, dx=g(x) \end{align}}$$ we get: $$\displaystyle\int ^b_{a}u\,dv=\Big[uv\Big]^b_{a}-\int ^{b}_{a}v \, du$$ ## Substitution in Definite Integrals (2.11) conditions: - $f$ is defined and continuous on $[a,b]$ - $\varphi$ is a function defined, continuous and continuously differentiable on an interval $J=[\alpha,\beta]$ or $J=[\beta,\alpha]$ - $\varphi(J)\subseteq[a,b]$ (that is, $f$ is continuous on $\varphi$'s image) - $\varphi(\alpha)=a$ and $\varphi(\beta)=b$ - Substitutions - (substitute $t=\varphi(x)$) $\displaystyle \int ^\beta_{\alpha}f(\varphi(t))\varphi'(x) \, dx =\int ^{\varphi(\beta)}_{\varphi(\alpha)}f(t) \, dt$ - (substitute $x=\varphi(t)$) $\displaystyle \int ^b_{a}f(x) \, dx =\int ^\beta_{\alpha}f(\varphi(t))\varphi'(t) \, dt$ > notation: others use in $u$ and $g$ instead of $t$ and $\varphi$