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

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Last Week: Two Techniques of IntegrationLast week we talked about two tools for integrating difficultfunctions:IIntegration by Substitution (Change of Variable)Zx=bx=af(u(x))dx=Zu=u(b)u=u(a)f(u)dxdudu=Zu=u(b)u=u(a)f(u)x0(u)du=Zu=u(b)u=u(a)f(u)u0(x)duIIntegration by PartsZudv=uv-ZvduZbaudv=uvba-Zbavdu
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)-1Zx2dx3
More TipsWe saw some instances where it was not obvious which tool(s) touse.ITrigonometric Identities: Eg.,IZtan(x)dx=Zsin(x)cos(x)dxIZsin2(x)dx=Z12-cos(2x)2dxIZsin3(x)dx=Z(1-cos2(x)) sin(x)dxwhich lead to a natural substitution.
More TipsIThe “·1” Trick , eg.Ztan-1(x)dx=Ztan-1(x)·1dx=xtan-1(x)-Zx·11 +x2dxby integration by parts, and the integralRx1+x2dxon the rightcan be solved with the substitutionu= 1 +x2.ICyclic Integration by Parts, eg.Zsin(x) cos(x)dx=sin2(x)-Zcos(x) sin(x)dx2Zsin(x) cos(x)dx=sin2(x) +CZsin(x) cos(x)dx=12sin2(x) +C
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Substitutions will not help you here! (At least, not directly.)This integral requires the method of “partial fractions”.
Trigonometric SubstitutionsThe name of the game is the following trigonometric identities:I1-sin2(θ) = cos2(θ)I1 + tan2(θ) =cos2(θ)cos2(θ)+sin2(θ)cos2(θ)= sec2(θ)Isec2(θ)-1 = tan2(θ)Notice that in each case, the right hand side of the identity is aperfect square.
Trigonometric Substitution—R11+x2dxRemember the identity1 + tan2(θ) = sec2(θ)Maybe we could use the substitutionx= tan(θ) to turn thedenominator 1 +x2into sec2(θ)? This will turn out to be usefulbecausedx= sec2(θ)dθ. Watch:Z11 +x2dx=Z11 + tan2(θ)·sec2(θ)dθ=Z1sec2(θ)sec2(θ)dθ=Zdθ=θ+C= tan-1(x) +C
Trigonometric Substitution—R101-x2dxWe already met this integral, and we know it gives the area ofone-quarter of the unit circle. This time, remember the identity1-sin2(θ)=cos2(θ)q1-sin2(θ)=|cos(θ)|So perhaps it makes sense to use the change of variablesx= sin(θ)dx= cos(θ)dθThis givesZx=1x=0p1-x2dx=Zθ=π/2θ=0q1-sin2(θ)·(cos(θ)dθ)=Zθ=π/2θ=0|cos(θ)| ·cos(θ)dθ=Zθ=π/2θ=0cos2(θ)dθsince cos(θ)0 for 0θπ/2.
Trigonometric Substitution—R101-x2dxSo a trigonometric substitutionx= sin(θ) shows thatZx=1x=0p1-x2

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