2-9 - Last Week Two Techniques of Integration Last week we talked about two tools for integrating difficult functions Integration by Substitution(Change

2-9 - Last Week Two Techniques of Integration Last week we...

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Last Week: Two Techniques of Integration Last week we talked about two tools for integrating difficult functions: I Integration by Substitution (Change of Variable) Z x = b x = a f ( u ( x )) dx = Z u = u ( b ) u = u ( a ) f ( u ) dx du du = Z u = u ( b ) u = u ( a ) f ( u ) x 0 ( u ) du = Z u = u ( b ) u = u ( a ) f ( u ) u 0 ( x ) du I Integration by Parts Z udv = uv - Z vdu Z b a udv = uv b a - Z b a vdu
Remember the following “tips” we talked about:IIf you see a compound function, think about trying asubstitution. Eg.,Z21e3xdx=Z63eu13du=13Z63euduIIf you’re integrating the product of a polynomial times afunction that anti-differentiates nicely (likeexor sin(x) orcos(x)), try integration by parts. Eg.,Zxexdx=xex-Zexdx=xex-ex+CIIf you have a polynomial times ln(x), ignore the last tip—differentiate the ln(x) and anti-differentiate the polynomial.Eg.,Zx2ln(x)=x33ln(x)-Zx331xdx=13x3ln(x)-1Zx2dx 3
More Tips We saw some instances where it was not obvious which tool(s) to use. I Trigonometric Identities: Eg., I Z tan( x ) dx = Z sin( x ) cos( x ) dx I Z sin 2 ( x ) dx = Z 1 2 - cos(2 x ) 2 dx I Z sin 3 ( x ) dx = Z (1 - cos 2 ( x )) sin( x ) dx which lead to a natural substitution.
More Tips I The “ · 1” Trick , eg. Z tan - 1 ( x ) dx = Z tan - 1 ( x ) · 1 dx = x tan - 1 ( x ) - Z x · 1 1 + x 2 dx by integration by parts, and the integral R x 1+ x 2 dx on the right can be solved with the substitution u = 1 + x 2 . I Cyclic Integration by Parts, eg. Z sin( x ) cos( x ) dx = sin 2 ( x ) - Z cos( x ) sin( x ) dx 2 Z sin( x ) cos( x ) dx = sin 2 ( x ) + C Z sin( x ) cos( x ) dx = 1 2 sin 2 ( x ) + C
More Techniques 0 Substitutions will not help you here! (At least, not directly.) This integral requires the method of “partial fractions”.
Trigonometric Substitutions The name of the game is the following trigonometric identities: I 1 - sin 2 ( θ ) = cos 2 ( θ ) I 1 + tan 2 ( θ ) = cos 2 ( θ ) cos 2 ( θ ) + sin 2 ( θ ) cos 2 ( θ ) = sec 2 ( θ ) I sec 2 ( θ ) - 1 = tan 2 ( θ ) Notice that in each case, the right hand side of the identity is a perfect square.

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