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, \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: $\int udv=uv-\int vdu$

The LIATE rule: choose $u=f(x)$ as the function that appears first in LIATE and choose $dv=g^{\prime }(x)dx$ 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

• $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln |f(x)|+C$ (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 $t={\phi }^{-1}(x)$

$\int f(x)dx=\int f(g(t))g^{\prime }(t)dt{|}_{t=g^{\prime }(x)}$

• (2.5) $f(\alpha x+\beta )$ form (where $\alpha \ne 0$)
  • $\int f(x)dx=F(x)+C⟹\int f(\alpha x+\beta )dx=\frac{1}{\alpha }F(\alpha x+\beta )+C$

Trigonometric substitution

Integrand containing	Substitution	Identity		Result of Substitution	After Simplification
$a^2-x^2$	$x=a\sin t$ $dx=a\cos tdt$ $t=\arcsin \frac{x}{a}$	${\sin }^{2}t+{\cos }^{2}t=1$	$-\frac{\pi }{2}\le t\le \frac{\pi }{2}$	$a^2(1-{\sin }^{2}t)$	$a^2{\cos }^{2}t$
$a^2+x^2$	$x=a\tan t$ $dx=a{\sec }^{2}tdt$ $t=\arctan \frac{x}{a}$	${\tan }^{2}t+1={\sec }^{2}t$	$-\frac{\pi }{2}<t<\frac{\pi }{2}$	$a^2(1+{\tan }^{2}t)$	$a^2{\sec }^{2}t$
$x^2-a^2$	x=a\sec t$$dx=a\sec t\tan t \,dt $t=\text{arcsec}\frac{x}{a}$	${\sec }^{2}t-1={\tan }^{2}t$	$0\le t<\frac{\pi }{2}$ or $\pi \le t<\frac{3\pi }{2}$	$a^2({\sec }^{2}t-1)$	$a^2{\tan }^{2}t$

Integrands	Substitutions
$a^2+(f(x){)}^2$	$f(x)=a\tan \theta$ or $a\cot \theta$ or $a\sinh \theta$
$a^2-(f(x){)}^2$	$f(x)=a\sin \theta$ or $a\cos \theta$
$(f(x){)}^2-a^2$	$f(x)=a\sec \theta$ or $a\csc$ or $a\cosh \theta$
$a+f(x)$	$f(x)=a{\tan }^2\theta$ or $a{\cot }^2\theta$ or $a{\sinh }^2\theta$
$a-f(x)$	$f(x)=a{\sin }^2\theta$ or $a{\cos }^2\theta$
$f(x)-a$	$f(x)=a{\sec }^2\theta$ or $a{\csc }^2\theta$ or ${\cosh }^2\theta$
$ax-x^2$	$x=a{\sin }^2\theta$ or $a{\cos }^{2}0$
$(a-x)(x-b)$	$x=a{\cos }^2\theta +b{\sin }^2\theta$
$\frac{a^n-x^n}{a^n+x^n}$	$x^n=a^n\cos \theta$

Tangent half-angle substitution

• $\sin x=\frac{2t}{1+t^2}$

• $\cos x=\frac{1-t^2}{1+t^2}$

• $dx=\frac{2}{1+t^2}dt$

• $\int f(\cos x)\sin xdx$ then we can substitute $t=\cos x$

• $\int f(\sin x)\cos xdx$ then we can substitute $t=\sin x$

• $\int {\sin }^{m}x{\cos }^{n}xdx$ ($m,n$ are nonnegative integers)

  • $m$ is odd:
    • $m=2k+1$
    • ${\sin }^{m}x={\sin }^{2k+1}x=({\sin }^{2}x{)}^k\sin x=(1-{\cos }^{2}x{)}^k\sin x$
  • $n$ is odd:
    • $n=2k+1$
    • ${\cos }^{n}x={\cos }^{2k+1}x=({\cos }^{2}x{)}^k\cos x=(1-{\sin }^{2}x{)}^k\cos x$
  • $m,n$ are even:
    • todo

Rational Functions

• $\frac{a}{(x-p{)}^k}$ (where $a,p\in \mathbb{R},k\in \mathbb{N}$)
  • $x-p=t$ and $dx=dt$
  • $\int \frac{a}{(x-p{)}^k}dx=\{\begin{array}{ll}a\ln |t|+C& k=1\\ a\frac{t^{1-k}}{1-k}+C& k\ge 2\end{array}=\{\begin{array}{ll}a\ln |x-p|+C& k=1\\ a\frac{(x-p{)}^{1-k}}{1-k}+C& k\ge 2\end{array}$
• $\frac{ax+b}{(x^2+px+q{)}^k}$ (where $a,b,p,q\in \mathbb{R},k\in \mathbb{N}$ and $x^2+px+q$ has no real roots)
  • Step 1:
    • $ax+b=\frac{a}{2}(2x+p)+b-\frac{ap}{2}$ (note $(2x+p)=(x^2+px+q{)}^{\prime }$)
    • $\int \frac{ax+b}{(x^2+px+q{)}^k}dx=\frac{a}{2}\int \frac{2x+p}{(x^2+px+q)}dx+\left(b-\frac{ap}{2}\right)\int \frac{dx}{(x^2+px+q{)}^k}$
  • Step 2: (evaluate the first integral)
    • $u=x^2+px+q$ and $du=(2x+p)dx$
    • $\int \frac{2x+p}{(x^2+px+q{)}^k}dx=\int \frac{du}{u^k}=\{\begin{array}{ll}\ln |u|+C& k=1\\ \frac{u^{1-k}}{1-k}+C& k\ge 2\end{array}$
    • $\int \frac{2x+p}{(x^2+px+q)}dx=\ln (x^2+px+q)+C$
    • $\int \frac{2x+p}{(x^2+px+q{)}^k}dx=\frac{1}{1-k}(x^2+px+q{)}^{1-k}+C$ (where $k\ge 2$)
  • Step 3: (evaluate the second integral)
    • $\int \frac{dx}{(x^2+px+q{)}^k}$
    • $(x^2+px+q)={\left(x+\frac{p}{2}\right)}^2+q-\frac{p^2}{4}$ (Completing the square)
    • $(x^2+px+q)={\left(x+\frac{p}{2}\right)}^2+d^2$
    • substitute: $x+\frac{p}{2}=t$ and $dx=dt$
    • $\int \frac{dx}{(x^2+px+q{)}^k}=\int \frac{dx}{{\left({\left(x+\frac{p}{2}\right)}^2+d^2\right)}^k}=\int \frac{dt}{(t^2+d^2{)}^k}$
    • $I_k=\int \frac{dt}{(t^2+d^2{)}^k}$ (see Examples e2.12)
    • $I_1=\frac{1}{d}\arctan \frac{t}{d}+C$
    • $I_{k+1}=\frac{1}{2k{d}^2}\cdot \frac{t}{(t^2+d^2{)}^k}+\frac{2k-1}{2k{d}^2}{I}_k$

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 improper (i.e. $\mathrm{deg}N\ge \mathrm{deg}D$), divide the denominator into the numerator to obtain $\frac{N(x)}{D(x)}=Q(x)+\frac{N_1(x)}{D(x)}$ where $\mathrm{deg}{N}_1(x)<\mathrm{deg}D(x)$. 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 $(a{x}^2+bx+c{)}^n$ where $a{x}^2+bx+c$ 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 $\frac{A_1}{(px+q)}+\frac{A_2}{(px+q{)}^2}+\cdots +\frac{A_m}{(px+q{)}^m}$
4. Quadratic factors: For each factor of the form $(a{x}^2+bx+c{)}^n$, the partial fraction decomposition must include the following sum of $n$ fractions. $\frac{B_{1}x+C_1}{a{x}^2+bx+c}+\frac{B_{2}x+C_2}{(a{x}^2+bx+c{)}^2}+\cdots +\frac{B_{n}x+C_n}{(a{x}^2+bx+c{)}^n}$

Definite Integrals

	Formula
Additivity	${\int }_a^{b}f(x)dx={\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx$	$a<c<b$ and $f$ is integrable on $[a,b]$
Shift Property	${\int }_a^{b}f(x)dx={\int }_{a-c}^{b-c}f(x+c)dx$
	${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$
	${\int }_a^b\alpha dx=\alpha (b-a)$
	${\int }_a^{a}f(x)dx=0$

• Cavalieri’s Quadrature Formula ${\int }_a^b{x}^{n}dx=\{\begin{array}{ll}\frac{b^{n+1}-a^{n+1}}{n+1}& n=1\\ \ln \left(\frac{b}{a}\right)& n=-1\end{array}$ (where $a,b,x\in \mathbb{R}$)

Integration by Parts

(2.10) Integration by Parts for Definite Integrals

${\int }_a^{b}f(x)g^{\prime }(x)dx=[f(x)g(x){]}_a^b-{\int }_a^b{f}^{\prime }(x)g(x)dx$

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:

${\int }_a^{b}udv=[uv{]}_a^b-{\int }_a^{b}vdu$

Substitution in Definite Integrals

(2.11) conditions:

• $f$ is defined and continuous on $[a,b]$
• $\phi$ is a function defined, continuous and continuously differentiable on an interval $J=[\alpha ,\beta ]$ or $J=[\beta ,\alpha ]$
• $\phi (J)\subseteq [a,b]$ (that is, $f$ is continuous on $\phi$‘s image)
• $\phi (\alpha )=a$ and $\phi (\beta )=b$
• Substitutions
  • (substitute $t=\phi (x)$) ${\int }_{\alpha }^{\beta }f(\phi (t)){\phi }^{\prime }(x)dx={\int }_{\phi (\alpha )}^{\phi (\beta )}f(t)dt$
  • (substitute $x=\phi (t)$) ${\int }_a^{b}f(x)dx={\int }_{\alpha }^{\beta }f(\phi (t)){\phi }^{\prime }(t)dt$

notation: others use in $u$ and $g$ instead of $t$ and $\phi$