# 2-4 - Last Time: Two Techniques of Integration Last time we...

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• CaptainStrawGoldfish10346
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Last Time: Two Techniques of IntegrationLast time 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
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WARNING: As we already saw in the case ofR204-x2dx,this “first instinct” substitution is not necessarily going towork!Zx=2x=0p4-x2dx=Zu=0u=4udxbut there’s no good way to change from “dx” to “du” without anextraxfactor.*Remember: Adding (or removing)constants(like-2) is easy—but you can’t add or remove functions!
Hidden Substitutions—Rtan(x)dxFirst, use a trigonometric identity to writeZtan(x)dx=Zsin(x)cos(x)dxNow look at what we have: the expression looks morecomplicated, but the sin(x) in the numerator is perfectly suited forthe substitutionu= cos(x), since this meansdu=-sin(x)dx.Ztan(x)dx=Zsin(x)cos(x)dx=-Z1cos(x)(-sin(x)dx)=-Z1udu=-ln|u|+C=-ln|cos(x)|+C=ln|sec(x)|+C

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