Math 231E. Fall 2013. Lecture 7A.Integration by PartsJust like u-substitution is “reverse chain rule”, we have the technique of integration by parts,which is “reverse product rule”.Recall thatddx(f(x)g(x)) = f0(x)g(x) + f(x)g0(x).Integrating both sides givesZbaddx(f(x)g(x)) dx =Zbaf0(x)g(x) dx +Zbaf(x)g0(x) dx,and the FTC gives us thatf(b)g(b) − f(a)g(a) =Zbaf0(x)g(x) dx +Zbaf(x)g0(x) dx,or, in indefinite form:Zf0(x)g(x) dx +Zf(x)g0(x) dx = f(x)g(x).If we writeu = f(x), v = g(x), du = f0(x) dx, dv = g0(x) dx,then we write this asZu dv +Zv du = uv,orZu dv = uv −Zv du.Good news: We got rid of the integral we wanted to solve!Bad news: And replaced it with another integral....So, of course, this may not make things better; in fact, making a wrong choice can make itworse!Example 31. Let us considerZxexdx.We want to choose u = x and dv = exdx. It is helpful to use such a table:u = x v = exdu = dx dv = exdxWe then haveZu dv = uv −Zv duZxexdx = xex−Zexdx= xex− ex+ C.Example 32. Now tryZx sin(x) dx.We want to chooseu = x v = −cos(x)du = dx dv = sin(x) dxWe then haveZu dv = uv −Zv duZx sin(x) dx = −x cos(x) −Z(−cos(x)) dx= −x cos(x) + sin(x) + C.Example 33. A tricky example. Let us considerZln x dx.It’s not clear what to do here because the function doesn’t really look like a product. But letus choose u = ln x and dv = dx, then we haveu = ln x v = xdu =dxxdv = dxWe then haveZu dv = uv −Zv duZln(x) dx = x ln(x) −Zdx= x ln(x) − x + C.Example 34.Zexsin(x) dx.We will choose u = sin(x) and dv = exdx, givingu = sin(x) v = exdu = cos(x) dx dv = exdxWe then haveZu dv = uv −Zv duZexsin(x) dx = exsin(x) −Zexcos(x) dx.Now we see that we have an integral that we still don’t know how to do, so let us tryintegration by parts again!To doZexcos(x) dx,We choose u = cos(x) and dv = exdx, givingu = cos(x) v = exdu = −sin(x) dx dv = exdxWe then haveZu dv = uv −Zv duZexcos(x) dx = excos(x) +Zexsin(x) dx.and it unfortunately looks as if we are back where we started. However, let us put it together:Zexsin(x) dx = exsin(x) −Zexcos(x) dx= exsin(x) −excos(x) +Zexsin(x) dx= exsin(x) − excos(x) −Zexsin(x) dx.We can put all of the integrals on the left-hand side to get2Zexsin(x) dx = exsin(x) − excos(x)Zexsin(x) dx =12(exsin(x) − excos(x)) .Example 35. Sometimes we need to do a little more work, but we will get there. Forexample, let us considerZsin5(x) dx.A reasonable guess might be u = sin4(x) and dv = sin(x) dx.u = sin4(x) v = −cos(x)du = 4 sin3(x) cos(x) dx dv = sin(x) dxWe then haveZu dv = uv −Zv duZsin5(x) dx = −sin4cos(x) −Z4 sin3(x) cos(x)(−cos(x)) dx= −sin4cos(x) +Z4 sin3(x) cos2(x) dx.This doesn’t necessarily look like an improvement, but of course we can use the fact thatcos2(x) = 1 − sin2(x),so the last line becomesZsin5(x) dx = −sin4cos(x) +Z4 sin3(x) cos2(x) dx= −sin4cos(x) +Z4 sin3(x)(1 − sin2(x)) dx= −sin4cos(x) +Z4 sin3(x) dx −Z4 sin5(x) dx.Again, collecting like terms on the left-hand side gives5Zsin5(x) dx = −sin4cos(x) +Z4 sin3(x) dx (4)Zsin5(x) dx = −15sin4cos(x) +45Zsin3(x) dx. (5)There is still some work to do, but the last integral that we obtained is better. So we tryZsin3(x) dxA reasonable guess might be u = sin2(x) and dv = sin(x) dx.u = sin2(x) v = −cos(x)du = 2 sin(x) cos(x) dx dv = sin(x) dxWe then haveZu dv = uv −Zv duZsin3(x) dx = −sin2cos(x) −Z2 sin(x) cos(x)(−cos(x)) dx= −sin3cos(x) +Z2 sin(x) cos2(x) dx.Using the same trick givesZsin3(x) dx = −sin3cos(x) +Z2 sin(x) cos2(x) dx= −sin3cos(x) + 2Zsin(x)(1 − sin2(x)) dx= −sin3cos(x) + 2Zsin(x) dx − 2Zsin3(x) dxand this givesZsin3(x) dx = −13sin3(x) cos(x) +23Zsin(x) dx = −13sin3(x) cos(x) −23cos(x).Adding this to (4) givesZsin5(x) dx = −15sin4cos(x) +45Zsin3(x) dx= −15sin4cos(x) +45−13sin3(x) cos(x) −23cos(x)= −15sin4cos(x) −415sin3(x) cos(x)
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