Refresh on Trig Identities, Integrals and Substitutions
19 Jun 2017 | Math CalculusLink to the lecture: Lec 27 MIT 18.01 Single Variable Calculus, Fall 2007
Trig Identities
Suppose we have a circle with radius 1 centered at the origin \((0,0)\). If we connect a point on the circle and the origin and form an angle \(\theta\) with the \(x\) axis, we have \(\sin^2(\theta)+\cos^2(\theta)=1\).
So then we have the double and half angle formulas:
\[\begin{array}{ccl} \cos(2\theta) &= &\cos^2(\theta)-\sin^2(\theta)\\ &=&\cos^2(\theta)-(1-\cos^2(\theta))\\ &=&2\cos^2(\theta)-1\\ \cos^2(\theta) &=& \frac{1+\cos(2\theta)}{2}\\ \sin^2(\theta) &=& \frac{1-\cos(2\theta)}{2}\\ \sin(2\theta)&=&2\sin(\theta)cos(\theta) \end{array}\]Trig Integrals
\(\int \sin^n(x)cos^m(x)dx\), \(m,n=0,1,2,...,\)
case 1: \(m=1\)
\[\int \sin^n(x)cos(x)dx\]substitute \(u=sin(x)\), \(du = cos(x)dx\)
\[\begin{array}{ccl} \int \sin^n(x)cos(x)dx &= &\int u^ndu\\ &=&\frac{u^{n+1}}{n+1}+C\\ &=&\frac{\sin^{n+1}(x)}{n+1}+C \end{array}\]case 2: \(m=2, n=3\)
\[\int \sin^3(x)cos^2(x)dx\]use \(\sin^2(x)=1-cos^2(x)\), substitute \(u=\cos(x)\), \(du=-\sin(x)dx\)
\[\begin{array}{ccl} \int \sin^3(x)cos^2(x)dx &= &\int (1-\cos^2(x))\sin(x)\cos^2(x)dx\\ &=&\int (\cos^2(x)-\cos^4(x))\sin(x)dx\\ &=&\int (u^2-u^4)(-du)dx\\ &=&-\frac{u^3}{3}+\frac{u^5}{5}+C\\ &=&-\frac{\cos^3(x)}{3}+\frac{\cos^5(x)}{5}+C \end{array}\]case 3: \(m=0, n=3\)
\[\int \sin^3(x)dx\]use \(\sin^2(x)=1-cos^2(x)\), substitute \(u=\cos(x)\), \(du=-\sin(x)dx\)
\[\begin{array}{ccl} \int \sin^3(x)dx &= &\int (1-\cos^2(x))\sin(x)dx\\ &=&\int (1-\cos^2(x))\sin(x)dx\\ &=&\int (1-u^2)(-du)dx\\ &=&-u+\frac{u^3}{3}+C\\ &=&-\cos(x)+\frac{\cos^3(x)}{3}+C \end{array}\]rule1: If we have some odd power to play with, we can use the double angle formula and substitution method to play around and find the integration.
case 4: \(m=2,n=0\)
\[\int \cos^2(x)dx\]use \(\cos^2(x)=\frac{1+cos(2x)}{2}\),
\[\begin{array}{ccl} \int \cos^2(x)dx &= &\int \frac{1+cos(2x)}{2}dx\\ &=&\frac{x}{2}+\frac{\sin(2x)}{4}+C \end{array}\]case 5: \(m=2,n=2\)
\[\int \cos^2(x)\sin^2(x)dx\]use \(\cos^2(x)=\frac{1+cos(2x)}{2},\sin^2(x)=\frac{1-cos(2x)}{2}\),
\[\begin{array}{ccl} \int \cos^2(x)\sin^2(x)dx &= &\int \frac{1+cos(2x)}{2}\cdot\frac{1-cos(2x)}{2}dx\\ &= &\int \frac{1-cos^2(2x)}{4}dx\\ &= &\int \frac{1}{4}-\frac{1+cos(4x)}{8}dx\\ &=& = \frac{x}{8}-\frac{\sin(4x)}{32} + C \end{array}\]Or we can use \(\sin^2(x)\cos^2(x) = (\sin(x)\cos(x))^2 = \frac{sin^2(2x)}{4}\)
rule 2: If we have even exponents, use the half angle identities to reduce the power of the trig functions.
Introduction to the polar coordinates
Any points \((x,y)\) on the circle \(x^2+y^2=a^2\) can be represented in the polar coordinates \((a\cos(\theta),a\sin(\theta))\). Sometimes it is useful to calculate the integrals of some function in the form \(f(y)=\sqrt{a^2-y^2}\). For example, we can substitute \(y=a\sin(\theta), dy=a\cos(\theta)d\theta\):
\[\begin{array}{ccl} \int \sqrt{a^2-y^2}dy &= &\int \sqrt{a^2-a\sin^2(\theta)}\cdot a\cos(\theta)d\theta\\ &= &\int a^2\cos^2(\theta)d\theta\\ &=& a^2(\frac{\theta}{2}+\frac{\sin(\theta)\cos(\theta)}{2}) + C\\ &=& \frac{a^2\arcsin(\frac{y}{a})}{2} + \frac{(a\sin(\theta))(a\cos(\theta))}{2} + C\\ &=& \frac{a^2\arcsin(\frac{y}{a})}{2} + \frac{y\sqrt{a^2-y^2}}{2} + C\\ \end{array}\]It becomes something easier to solve.
Other Trig Identities and Integration
\[\begin{array}{cl}\sec(x)=\frac{1}{\cos(x)}&\csc=\frac{1}{\sin(x)}\\ \tan(x)=\frac{\sin(x)}{\cos(x)}&\cot(x) = \frac{\cos(x)}{\sin(x)} \end{array}\] \[\sec^2(x) = 1+\tan^2(x)\] \[\begin{array}{cl}\frac{d}{dx}\tan(x)=\sec^2(x)&\frac{d}{dx}\sec(x)=\sec(x)\tan(x)\\ \int\tan(x)dx=-\ln\vert\cos(x)\vert+c&\int\sec(x)dx = \ln(\sec(x)+\tan(x))+c \end{array}\]Example 1:
using \(u=tan(x),du=\sec^2(x)dx\)
\[\begin{array}{ccl} \int\sec^4(x)dx&=&\int(1+\tan^2(x))\sec^2(x)dx\\ &=&\int(1+u^2)du\\ &=&u+\frac{u^3}{3}+c\\ &=&\tan(x)+\frac{\tan^3(x)}{3}+c \end{array}\]Example 2:
using \(x=tan(\theta),dx=\sec^2(\theta)d\theta,u=\sin(\theta),du=\cos(\theta)d\theta\)
\[\begin{array}{ccl} \int\frac{dx}{x^2\sqrt{1+x^2}}&=&\int\frac{\sec^2(\theta)d\theta}{\tan^2(x)\sec(\theta)}\\ &=&\int\frac{\cos{\theta}}{\sin^2(\theta)}d\theta\\ &=&\int\frac{du}{u^2}\\ &=&-\frac{1}{u}+c\\ &=&-\frac{1}{\sin(\theta)}+c\\ &=&-\csc(\theta)+c\\ &=&-\csc(\arctan(x))+c \end{array}\]rule 3, substitute the following square roots using trigs:
\[\begin{array}{ccl}\sqrt{a^2-x^2}&\to&x=a\sin(\theta)/a\cos(\theta)\\ \sqrt{a^2+x^2}&\to&x=a\tan(\theta)\\ \sqrt{x^2-a^2}&\to&x=a\sec(\theta) \end{array}\]These substitutions can be derived using a right triangle by assigning the 1 and x to the hpp, opp and adj.
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