Trig+Substitution

Recall:

Trig Identities: math 1.) sin^2x + cos^2x = 1 math math 2) */cos^2x: tan^2x +1 = sec^2x math math 3) */sin^2x: 1 + cot^2x= csc^2x math

Trig Substiution Rules

$\sqrt{x^2+a^2}$ math || math $x=atanx$ math || math $tan^2x+1=sec^2x$ math || $\sqrt{a^2-x^2}$ math || math x=asinx$ math || math $1-sin^2x=cos^2x$ math || $\sqrt{x^2-a^2}$ math || math $x=asecx$ math || math $sec^2x-1=tan^2x$ math ||
 * Expression || Substitution || Identity ||
 * math
 * math
 * math

Examples:

math 1.) \int \frac{1}{\sqrt{1-x^2}}dx math math use: asinx dx=cosxdx \sqrt{1-x^2}=>\sqrt{1-sin^2x}=\sqrt{cos^2x}=cosx math math \int \frac{cosxdx}{cosx}=\int dx= x + c math

Plug back in "x"

math x+c = \sin^{-1}(x)+c math

Therefore we know

math \int \frac{1}{\sqrt{1-x^2}}dx=\arcsin(x)+c math

The general form of this is:

math \int \frac{1}{\sqrt{a^2-x^2}}dx=\arcsin (\frac{x}{a})+c math

math 2.)\int \frac{\sqrt{9-x^2}}{x^2}dx math

math use: x=3sinu => dx=3cosu math

math x^2=9sin^2u math

math \sqrt{9-x^2}=\sqrt{9-9sin^2}=\sqrt{9cos^2}=3cosu math

math =\int \frac{3cosu}{9sin^2u}*3cosudu math

math =\int \frac{cos^2u}{sin^2u}du math

math math
 * Recall: cos^2x+sin^2x=1 => cos^2x=1-sin^2x

math =\int \frac{1-sin^2u}{sin^2u}du math

math =\int \csc ^2udu-1\int du math

math =-\cot(u)-u+c math

Plug back in "x"

math Recall: x=3\sin u math

math \frac{x}{3}=sinu math math u=\arcsin\frac{x}{3} math

Therefore: math \cot(u)=\frac{adjacent}{opposite}=\frac{\sqrt{9-x^2}}{x} math

math =-\frac{\sqrt{9-x^2}}{x}-\arcsin \frac{x}{3}+c math

Good outside websites:
 * http://www.sosmath.com/calculus/integration/trigsub/trigsub.html
 * http://www.math.hmc.edu/calculus/tutorials/trig_substitution/