1. Direct Integration
It's not too difficult to recognise integrals of the form $\int f(x)f'(x)dx=\int f(x) d[f(x)]$.
Example:
It's not too difficult to recognise integrals of the form $\int f(x)f'(x)dx=\int f(x) d[f(x)]$.
Example:
$\int (5x^3+4x^2)(15x^2+8x)dx\\= \int (5x^3+4x^2) d(5x^3+4x^2)\\= \frac{(5x^3+4x^2)^2}{2}+C$
2. Expansion
Whenever we see (x ± )^(a number) and it seems that substitution won't work, expansion is a good approach.
Example:
$\int (x^2-1)^6 dx\\= \int (x^{12}-6x^{10}+15x^8-20x^6+15x^4-6x^2+1) dx\\= \frac{x^{13}}{13}-6\frac{x^{11}}{11}+5\frac{x^9}{3}-20\frac{x^7}{7}+\frac{x^5}{3}-2x^3+x+C$
Whenever we see (x ± )^(a number) and it seems that substitution won't work, expansion is a good approach.
Example:
$\int (x^2-1)^6 dx\\= \int (x^{12}-6x^{10}+15x^8-20x^6+15x^4-6x^2+1) dx\\= \frac{x^{13}}{13}-6\frac{x^{11}}{11}+5\frac{x^9}{3}-20\frac{x^7}{7}+\frac{x^5}{3}-2x^3+x+C$
3. Substitution
Fundamental Theorem of Calculus
Example:
$\large \int_0^1 \frac{x^t-1}{\ln x} dx$ where $t\geq 0$
Let $\large{F(t)=\int_0^1 \frac{x^t-1}{\ln x}dx\\
\begin{align}F'(t)&=\frac{d}{dt}\int_0^1\frac{x^t-1}{\ln x}dx\\
&=\int_0^1 \frac{d}{dt}\frac{x^t-1}{\ln x}dx\\
&=\int_0^1\frac{1}{\ln x}\frac{d}{dt}(x^t-1)dx \:\:\:\:\:[\frac{d}{dx}x^t=x^t \ln x]\\
&=\int_0^1 x^t dx\\
&=\frac{1}{t+1}\end{align}}$
Integrate again,
$F(t)=\ln(t+1)+C$
Since $F(0)=0$, we have $C=0$ and $F(t)=\ln(t+1)$.
Make use of known integrals
i. u-substitution
ii. trig-substitution
[How to find the right substitution?]
[<=> change of coordinates]
[<=> linear algebra]
[<=> change of coordinates]
[<=> linear algebra]
4. Integration by parts
$\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\\ uv=\int udv+\int vdu\\ \int udv=uv - \int vdu$
[How to know which is u and which is v?]
[How to know which is u and which is v?]
6. Reduction formula
Technical.
Technical.
Special methods:
3. Substitution (cont'd)
iii. Weierstrass t-substitution / Tangent half-angle substitution
Fundamental Theorem of Calculus
Example:
$\large \int_0^1 \frac{x^t-1}{\ln x} dx$ where $t\geq 0$
Let $\large{F(t)=\int_0^1 \frac{x^t-1}{\ln x}dx\\
\begin{align}F'(t)&=\frac{d}{dt}\int_0^1\frac{x^t-1}{\ln x}dx\\
&=\int_0^1 \frac{d}{dt}\frac{x^t-1}{\ln x}dx\\
&=\int_0^1\frac{1}{\ln x}\frac{d}{dt}(x^t-1)dx \:\:\:\:\:[\frac{d}{dx}x^t=x^t \ln x]\\
&=\int_0^1 x^t dx\\
&=\frac{1}{t+1}\end{align}}$
Integrate again,
$F(t)=\ln(t+1)+C$
Since $F(0)=0$, we have $C=0$ and $F(t)=\ln(t+1)$.
Make use of known integrals
Manipulations
Change of coordinates (Transform into a double integral then switch to polar coordinates)
Let I = $\int_0^\infty e^{-x^2}dx$
$I^2=\int_0^\infty e^{-x^2}dx \int_0^\infty e^{-y^2}dy$
$=\int_0^\infty[\int_0^\infty e^{-x^2}dx]e^{-y^2}dy$
$=\int_0^\infty \int_0^\infty e^{-(x^2+y^2)}dx dy$
$=\int_0^{\frac{\pi}{2}}\int_0^\infty e^{-r^2}rdr d\theta$
$=\int_0^{\frac{\pi}{2}}-\frac{1}{2}e^{-r^2}|_0^\infty d\theta$
$=\frac{\pi}{4}$
It follows that I = $\frac{\sqrt\pi}{2}$.
Evaluate a general integral first: Parameter differentiation p56
Symmetry
http://www2.math.umd.edu/~punshs/Calculus/Integration.pdf
Odd function
$\int_{-2}^2 (1-2x-x^{16} \sin x+17x^7\sqrt{1+x^2}-x^{11} \cos x )dx$
Let $f(x)=-2x-x^{16} \sin x+17x^7\sqrt{1+x^2}-x^{11} \cos x$
$f(-x)=2x+x^{16} \sin x-17x^7\sqrt{1+x^2}+x^{11} \cos x=-f(x)$
$\Rightarrow \int_{-2}^2 (1-2x-x^{16} \sin x+17x^7\sqrt{1+x^2}-x^{11} \cos x )dx\\=\int_{-2}^2 (1+f(x))dx\\=4$
Linear Algebra
i. Orthogonality
Example:
$\int_{-\pi}^\pi (3+2\sin x+3\cos x)(1+4\sin x)dx$
$1,\sin x,\cos x$ are orthogonal on $[-\pi,\pi]$
$=\int_{-\pi}^\pi 3 dx+\int_{-\pi}^\pi 8\sin^2 x dx$
$=3(2\pi)+8\pi$
$=14\pi$
ii. Change of basis
Example:
Let $\mathbb{B}=\left\{1,\cos t,\cos^2 t,...,\cos^6 t\right\}$ and $\mathbb{C}=\left\{1,\cos t,\cos 2t,...,\cos 6t\right\}$.
By De Moivre's theorem,
$\cos 2t=-1+2\cos^2 t$
$\cos 3t=-3\cos t+4\cos^3 t$
$\cos 4t=1-8\cos^2 t+8\cos^4 t$
$\cos 5t=5\cos t-20\cos^3 t+16\cos^5 t$
$\cos 6t=-1+18\cos^2 t-48\cos^4 t+32\cos^6 t$
$P=[\mathbb{B}]_\mathbb{C}=\begin{pmatrix} 1 & 0 & -1 & 0 & 1 & 0 & -1\\ 0 & 1 & 0 & -3 & 0 & 5 & 0 \\ 0 & 0 & 2 & 0 & -8 & 0 & 18 \\ 0 & 0 & 0 & 4 & 0 & -20 & 0 \\ 0 & 0 & 0 & 0 & 8 & 0 & -48 \\ 0 & 0 & 0 & 0 & 0 & 16 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 32 \end{pmatrix}$
$P^{-1}=\frac{1}{32}\begin{pmatrix} 32 & 0 & 16 & 0 & 12 & 0 & 10\\ 0 & 32 & 0 & 24 & 0 & 20 & 0 \\ 0 & 0 & 16 & 0 & 16 & 0 & 15 \\ 0 & 0 & 0 & 8 & 0 & 10 & 0 \\ 0 & 0 & 0 & 0 & 4 & 0 & 6 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}$
Say we want to evaluate $\int (5\cos^3 t-6\cos^4 t+5\cos^5 t-12\cos^6 t)dt$.
$P^{-1}(0,0,0,5,-6,5,-12)=(-6,\frac{55}{8},-\frac{69}{8},\frac{45}{16},-3,\frac{5}{16},\frac{-3}{8})$
$\int (5\cos^3 t-6\cos^4 t+5\cos^5 t-12\cos^6 t)dt\\
=\int (-6+\frac{55}{8}\cos t-\frac{69}{8}\cos 2t+\frac{45}{16}\cos 3t-3\cos 4t+\frac{5}{16}\cos 5t-\frac{3}{8}\cos 6t)dt\\
=-6t+\frac{55}{8}\sin t-\frac{69}{16}\sin 2t+\frac{15}{16}\sin 3t-\frac{3}{4}\sin 4t+\frac{1}{16}\sin 5t-\frac{1}{16}\sin 6t+C$
Geometry
$\int_0^1 [(1-x^5)^{\frac{1}{4}}-(1-x^4)^{\frac{1}{5}}]dx$
Note that $\int_0^1 (1-x^5)^{\frac{1}{4}}dx=\int_0^1 (1-y^4)^{\frac{1}{5}}dy$.
Both are the area bounded by $x^5+y^4=1$ with the x and y-axes.
So answer: 0.
Recursive [prob solv] refer to notebook
Probability density function
This will be explained in another post.
Examples:
Normal distribution
$\large \int_{-\infty}^\infty e^{-4x^2}dx$
$\large =\int_{-\infty}^\infty exp\left\{\frac{x^2}{\frac{1}{4}} \right\}dx$
$\large =\int_{-\infty}^\infty exp\left\{\frac{x^2}{2(\frac{1}{8})} \right\}dx$
$\large \sigma^2=\frac{1}{8}$ and $\mu =0$
$\large{\begin{align}\therefore \int_{-\infty}^\infty e^{-4x^2}dx &= \sqrt{\frac{1}{8}} \sqrt{2\pi} \int_{-\infty}^\infty \frac{1}{\sqrt{\frac{1}{8}} \sqrt{2\pi}}exp\left\{\frac{x^2}{2(\frac{1}{8})} \right\}dx\\
&=\sqrt{\frac{1}{8}} \sqrt{2\pi}\\
&=\frac{\sqrt{\pi}}{2}\end{align}}$
Gamma distribution
$\large \int_0^\infty 81e^{-3x}x^4dx$
$\large =\frac{\Gamma(5)}{3}\int_0^\infty \frac{3e^{-3x}(3x)^4}{\Gamma(5)}dx$
$\large =\frac{4!}{3}$
$\large =8$
We can integrate a function of the form $e^{-kx}x^s$ using gamma distribution. This avoids integrating by parts as many as s times.
Beta distribution
$\large \int_0^1 x^5 (1-x^2)^9 dx$
$\large =\int_0^1 (x^2)^{\frac{5}{2}} (1-x^2)^9 dx$
$\large \stackrel{y=x^2}{=}\int_0^1 \frac{y^{\frac{5}{2}}(1-y)^9}{2\sqrt{y}}dy$
$\large =\frac{1}{2}\int_0^1 y^2(1-y)^9dy$
$\large =\frac{1}{2}B(3,10) \int_0^1 \frac{y^2(1-y)^9}{B(3,10)}dy$
$\large =\frac{1}{2}\frac{\Gamma(3)\Gamma(10)}{\Gamma(13)}$
$\large =\frac{1}{2}\frac{2!9!}{12!}$
$\large =\frac{1}{12\cdot 11\cdot 10}$
$\large =\frac{1}{1320}$
Tabular Integration by Parts
Refer to http://www.maa.org/sites/default/files/pdf/mathdl/CMJ/Horowitz307-311.pdf
Comparing coefficients
Refer to https://johnmayhk.wordpress.com/2014/08/08/integration-by-comparing-coefficients/
Laplace transform <=> residue theory
Complex analysis
Change of coordinates (Transform into a double integral then switch to polar coordinates)
Let I = $\int_0^\infty e^{-x^2}dx$
$I^2=\int_0^\infty e^{-x^2}dx \int_0^\infty e^{-y^2}dy$
$=\int_0^\infty[\int_0^\infty e^{-x^2}dx]e^{-y^2}dy$
$=\int_0^\infty \int_0^\infty e^{-(x^2+y^2)}dx dy$
$=\int_0^{\frac{\pi}{2}}\int_0^\infty e^{-r^2}rdr d\theta$
$=\int_0^{\frac{\pi}{2}}-\frac{1}{2}e^{-r^2}|_0^\infty d\theta$
$=\frac{\pi}{4}$
It follows that I = $\frac{\sqrt\pi}{2}$.
Evaluate a general integral first: Parameter differentiation p56
Symmetry
http://www2.math.umd.edu/~punshs/Calculus/Integration.pdf
Odd function
$\int_{-2}^2 (1-2x-x^{16} \sin x+17x^7\sqrt{1+x^2}-x^{11} \cos x )dx$
Let $f(x)=-2x-x^{16} \sin x+17x^7\sqrt{1+x^2}-x^{11} \cos x$
$f(-x)=2x+x^{16} \sin x-17x^7\sqrt{1+x^2}+x^{11} \cos x=-f(x)$
$\Rightarrow \int_{-2}^2 (1-2x-x^{16} \sin x+17x^7\sqrt{1+x^2}-x^{11} \cos x )dx\\=\int_{-2}^2 (1+f(x))dx\\=4$
Linear Algebra
i. Orthogonality
Example:
$\int_{-\pi}^\pi (3+2\sin x+3\cos x)(1+4\sin x)dx$
$1,\sin x,\cos x$ are orthogonal on $[-\pi,\pi]$
$=\int_{-\pi}^\pi 3 dx+\int_{-\pi}^\pi 8\sin^2 x dx$
$=3(2\pi)+8\pi$
$=14\pi$
ii. Change of basis
Example:
Let $\mathbb{B}=\left\{1,\cos t,\cos^2 t,...,\cos^6 t\right\}$ and $\mathbb{C}=\left\{1,\cos t,\cos 2t,...,\cos 6t\right\}$.
By De Moivre's theorem,
$\cos 2t=-1+2\cos^2 t$
$\cos 3t=-3\cos t+4\cos^3 t$
$\cos 4t=1-8\cos^2 t+8\cos^4 t$
$\cos 5t=5\cos t-20\cos^3 t+16\cos^5 t$
$\cos 6t=-1+18\cos^2 t-48\cos^4 t+32\cos^6 t$
$P=[\mathbb{B}]_\mathbb{C}=\begin{pmatrix} 1 & 0 & -1 & 0 & 1 & 0 & -1\\ 0 & 1 & 0 & -3 & 0 & 5 & 0 \\ 0 & 0 & 2 & 0 & -8 & 0 & 18 \\ 0 & 0 & 0 & 4 & 0 & -20 & 0 \\ 0 & 0 & 0 & 0 & 8 & 0 & -48 \\ 0 & 0 & 0 & 0 & 0 & 16 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 32 \end{pmatrix}$
$P^{-1}=\frac{1}{32}\begin{pmatrix} 32 & 0 & 16 & 0 & 12 & 0 & 10\\ 0 & 32 & 0 & 24 & 0 & 20 & 0 \\ 0 & 0 & 16 & 0 & 16 & 0 & 15 \\ 0 & 0 & 0 & 8 & 0 & 10 & 0 \\ 0 & 0 & 0 & 0 & 4 & 0 & 6 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}$
Say we want to evaluate $\int (5\cos^3 t-6\cos^4 t+5\cos^5 t-12\cos^6 t)dt$.
$P^{-1}(0,0,0,5,-6,5,-12)=(-6,\frac{55}{8},-\frac{69}{8},\frac{45}{16},-3,\frac{5}{16},\frac{-3}{8})$
$\int (5\cos^3 t-6\cos^4 t+5\cos^5 t-12\cos^6 t)dt\\
=\int (-6+\frac{55}{8}\cos t-\frac{69}{8}\cos 2t+\frac{45}{16}\cos 3t-3\cos 4t+\frac{5}{16}\cos 5t-\frac{3}{8}\cos 6t)dt\\
=-6t+\frac{55}{8}\sin t-\frac{69}{16}\sin 2t+\frac{15}{16}\sin 3t-\frac{3}{4}\sin 4t+\frac{1}{16}\sin 5t-\frac{1}{16}\sin 6t+C$
Geometry
$\int_0^1 [(1-x^5)^{\frac{1}{4}}-(1-x^4)^{\frac{1}{5}}]dx$
Note that $\int_0^1 (1-x^5)^{\frac{1}{4}}dx=\int_0^1 (1-y^4)^{\frac{1}{5}}dy$.
Both are the area bounded by $x^5+y^4=1$ with the x and y-axes.
So answer: 0.
Recursive [prob solv] refer to notebook
Probability density function
This will be explained in another post.
Examples:
Normal distribution
$\large \int_{-\infty}^\infty e^{-4x^2}dx$
$\large =\int_{-\infty}^\infty exp\left\{\frac{x^2}{\frac{1}{4}} \right\}dx$
$\large =\int_{-\infty}^\infty exp\left\{\frac{x^2}{2(\frac{1}{8})} \right\}dx$
$\large \sigma^2=\frac{1}{8}$ and $\mu =0$
$\large{\begin{align}\therefore \int_{-\infty}^\infty e^{-4x^2}dx &= \sqrt{\frac{1}{8}} \sqrt{2\pi} \int_{-\infty}^\infty \frac{1}{\sqrt{\frac{1}{8}} \sqrt{2\pi}}exp\left\{\frac{x^2}{2(\frac{1}{8})} \right\}dx\\
&=\sqrt{\frac{1}{8}} \sqrt{2\pi}\\
&=\frac{\sqrt{\pi}}{2}\end{align}}$
Gamma distribution
$\large \int_0^\infty 81e^{-3x}x^4dx$
$\large =\frac{\Gamma(5)}{3}\int_0^\infty \frac{3e^{-3x}(3x)^4}{\Gamma(5)}dx$
$\large =\frac{4!}{3}$
$\large =8$
We can integrate a function of the form $e^{-kx}x^s$ using gamma distribution. This avoids integrating by parts as many as s times.
Beta distribution
$\large \int_0^1 x^5 (1-x^2)^9 dx$
$\large =\int_0^1 (x^2)^{\frac{5}{2}} (1-x^2)^9 dx$
$\large \stackrel{y=x^2}{=}\int_0^1 \frac{y^{\frac{5}{2}}(1-y)^9}{2\sqrt{y}}dy$
$\large =\frac{1}{2}\int_0^1 y^2(1-y)^9dy$
$\large =\frac{1}{2}B(3,10) \int_0^1 \frac{y^2(1-y)^9}{B(3,10)}dy$
$\large =\frac{1}{2}\frac{\Gamma(3)\Gamma(10)}{\Gamma(13)}$
$\large =\frac{1}{2}\frac{2!9!}{12!}$
$\large =\frac{1}{12\cdot 11\cdot 10}$
$\large =\frac{1}{1320}$
Tabular Integration by Parts
Refer to http://www.maa.org/sites/default/files/pdf/mathdl/CMJ/Horowitz307-311.pdf
Comparing coefficients
Refer to https://johnmayhk.wordpress.com/2014/08/08/integration-by-comparing-coefficients/
Laplace transform <=> residue theory
Complex analysis
Good!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
ReplyDelete