Notes for Thurs 8 Sept Calculus II Fall 2005 New York University Instructor: Tyler Neylon Scribe: Kelsey Williams

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1 Notes for Thurs 8 Sept Clculus II Fll 00 New York University Instructor: Tyler Neylon Scribe: Kelsey Willims 8. Integrtion by Prts This section is primrily bout the formul u dv = uv v ( ) which is essentilly the integrl version of the proct rule. Theorem (Fundmentl Theorem of Clculus (prt of it, nywy)) b f (x)dx = f(b) f() As quick exmple, if f(0) = 0 then y 0 f (x)dx = f(y). So now let s prove tht integrtion by prts relly works. Strt with the proct rule: (f g) = f g + fg nd pply b dx to both sides. We get b (fg) dx = b (f g + fg )dx ( ) The left-hnd side is just [ fg ] b, which is specil nottion for f(b)g(b) f()g(). The right-hnd side we cn split into the sum of two integrls, s b f g dx + b fg dx. So relly eqution ( ) is the sme s [ ] b b b fg = f g dx + fg dx which, moving the terms round, is equivlent to b [ ] b b fg dx = fg f g dx, which is just nother form of the stndrd integrtion by prts formul ( ) to see this, just set u = f(x) nd v = g(x), so tht = f (x)dx nd dv = g (x)dx. Exmple Find xe x dx. Let u = x = dx dv = e x dx v = e x so tht xe x dx = xe x e x dx = xe x e x + C.

2 Exmple Find x e x dx. Let u = x = x dx so tht our integrl becomes x e x + dv = e x dx v = e x xe x dx. Well, we cn t just integrte the lst bit directly, but it does look simpler thn it ws before, so let s try ye olde integrtion by prts one more time: u = x = dx dv = e x dx v = e x nd our expression for the originl integrl is now = x e x xe x + e x dx = x e x xe x e x + C. Exmple 3 Find x cos(x 3 ) dx. In this exmple, the trick is to pick dv which is esy to integrte. If we just pick dv = cos(x 3 ), then how do we find v? It s not esy. But the following choice, mde specificlly so tht v is findble, works well: The integrl is now u = x 3 dv = x cos(x 3 ) dx = 3x dx v = 3 sin(x3 ). 3 x3 sin(x 3 ) x sin(x 3 ) dx. Notice tht the integrnd looks much simpler. At this point we cn finish off by using substitution: u = x 3 = 3x dx. Then we hve 3 x3 sin(x 3 ) 3 sin(u) = 3 ( x 3 sin(x 3 ) + cos(x 3 ) ) + C. Exmple 4 Find ln(x) dx. In this cse, it might be tricky to see how to split up the integrnd since there isn t even multipliction in there to split up! However, there s nothing bout integrtion by prts which sys tht u or dv must be

3 obvious prts of proct in the integrnd. This is one exmple where we hve to do something tht t first might seem little nonobvious, which is setting dv = dx. This is perfectly legl, since relly ln(x) = ln(x). In generl, we could hve even chosen something crzy like dv = e x dx nd then sid u = e x ln(x). Integrtion by prts still works! But tht lst crzy choice of u nd dv wouldn t help us solve the originl integrl it would just mke things more complex. So we ll proceed with u = ln(x) dv = dx = dx/x v = x. From this we see tht ln(x) dx = x ln(x) x dx = x ln(x) x + C. x 8.6 Rtionlizing Substitutions (k Dirty Tricks ) This section is ll bout turning some complicted-looking integrnds into wht is clled rtionl function, for which there is fixed set of rules to integrte. A rtionl function is function f(x) of the form f(x) = polynomil polynomil. So x /( x) is rtionl function, but x is not. Any polynomil still counts s rtionl function, since we could just think of the denomintor s being = (nd the number by itself still counts s polynomil). Once we ve used substitution to turn n integrnd into rtionl function, here is the strtegy to use:. If you see immeditely how to solve it, do so; otherwise go on:. Perform long division on the frction of polynomils 3. Find prtil frction decomposition of the remining frction (if there is one) If you follow these steps correctly, you should be ble to integrte ny rtionl function whtsoever. But if you follow them incorrectly... well, tht s not s good. For more bout how to integrte rtionl functions, check out 8. in the book. We ll see how to do this turning-into--rtionl-function by some illustrtive exmples. 3

4 Exmple Find dx +. In cses like this, it is often good ide to mke substitution x of the form u m = x, where m is some common multiple of the level of roots in the integrnd. So in this cse we ll use u = x u = x u = dx, which turns the problem into ( u + u = ). + u We cn rrive t tht lst equlity by using long division. From here we cn finish: = + u = u ln + u + C = x ln + x + C. Exmple 6 Find dx 3 x+ x. In this cse, we hve third root nd squre ( second ) root, so the lest common multiple between these is six: u 6 = x 6u = dx gives us the new integrl 6u u + u 3 = 6 u 3 + u. Now do some long division, ( = 6 u u + ) + u ( = 6 3 u3 ) u + u ln + u + C. Finlly plug bck in 6 x for u: = x 3 3 x x 6 ln + 6 x + C. Now it s time to explore the wonderful world of trigonometric functions nd the substitutions who help them. Bsiclly, there s only one substitution we ll study here. The relly cool prt is tht this single substitution cn turn ny function built out of sin, cos, tn, nd the 4 bsic opertors (+,,, /) into rtionl function which we cn then integrte. Let s do running exmple to see how everything works out. We ll try to integrte dx 3 sin x 4 cos x 4

5 The substitution itself is u = tn(x/). Solve for x to see tht x = tn (u) so tht dx = /( + u ). Ok, so fr so good. But now we hve something little complex-looking (nd not very rtionl-function-looking): +u 3 sin( tn (u)) 4 cos( tn (u)) Ick! It does not pper tht this substitutuion ws ny good t ll. But wit! Actully, we cn relly simplify the sin nd cos bits. Let s notice few trig equlities: cos(x/) = sec(x/) = tn (x/) + = + u nd sin(x/) = cos(x/) tn(x/) = u + u. From here we cn use the double-ngle formule sin(θ) = sin(θ) cos(θ) nd cos(θ) = cos (θ) sin (θ) with θ = x/ so see tht sin(x) = u + u nd cos(x) = u + u. Now we re in position to relly simplify our integrl: ( dx 3 sin x 4 cos x = + u + u 6u 4 + 4u = u + 3u = (u )(u + ) = ) u u +. To find tht lst expression, we used prtil frction decomposition. Almost done: = ln u ln u + + C Good work! = ln u u + + C = ln tn(x/) tn(x/) + + C.

6 As summry of this type of substitution, here is tble of the importnt equlities to remember: u = tn ( ) x x = tn dx = (u) +u sin(x) = u +u cos(x) = u +u 6

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