Lecture 30 18.01 Fall 2006 Lecture 30: Integration by Parts, Reduction Formulae Integration by Parts Remember the product rule: (uv)� = u� v + uv � We can rewrite that as uv � = (uv)� − u� v Integrate this to get the formula for integration by parts: � � � uv dx = uv − u� v dx � Example 1. tan−1 x dx. At ﬁrst, it’s not clear how integration by parts helps. Write � � � tan−1 x dx = tan−1 x(1 · dx) = uv � dx with u = tan−1 x and v � = 1. Therefore, 1 1 + x2 Plug all of these into the formula for integration by parts to get: � � � tan−1 x dx = uv � dx = (tan−1 x)x − v = x and u� = 1 (x)dx 1 + x2 1 ln |1 + x2 | + c 2 = x tan−1 x − Alternative Approach to Integration by Parts As above, the product rule: (uv)� = u� v + uv � can be rewritten as uv � = (uv)� − u� v This time, let’s take the deﬁnite integral: � b � b � � u� v dx (uv) dx − uv dx = a b � a a 1 Lecture 30 18.01 Fall 2006 By the fundamental theorem of calculus, we can say � b �b � � uv � dx = uv� − a a b u� v dx a Another notation in the indeﬁnite case is � � u dv = uv − v du This is the same because dv = v � dx =⇒ uv � dx = u dv and du = u� dx =⇒ u� v dx = vu� dx = v du � (ln x)dx Example 2. 1 u = ln x; du = dx and dv = dx; v = x x � � � � � 1 (ln x)dx = x ln x − x dx = x ln x − dx = x ln x − x + c x We can also use “advanced guessing” to solve this problem. We know that the derivative of something equals ln x: d (??) = ln x dx Let’s try d 1 (x ln x) = ln x + x · = ln x + 1 dx x That’s almost it, but not quite. Let’s repair this guess to get: d (x ln x − x) = ln x + 1 − 1 = ln x dx Reduction Formulas (Recurrence Formulas) � Example 3. n (ln x) dx Let’s try: u = (ln x)n =⇒ u� = n(ln x)n−1 � � 1 x v � = dx; v = x Plugging these into the formula for integration by parts gives us: � n � n (ln x) dx = x(ln x) − � � 1 � 1� x � dx � x n−1 n(ln x) Keep repeating integration by parts to get the full formula: n → (n − 1) → (n − 2) → (n − 3) → etc � Example 4. xn ex dx Let’s try: u = xn =⇒ u� = nxn−1 ; 2 v � = ex =⇒ v = ex Lecture 30 18.01 Fall 2006 Putting these into the integration by parts formula gives us: � � xn ex dx = xn ex − nxn−1 ex dx Repeat, going from n → (n − 1) → (n − 2) → etc. Bad news: If you change the integrals just a little bit, they become impossible to evaluate: � � −1 �2 tan x dx = impossible � ex dx = also impossible x Good news: When you can’t evaluate an integral, then � 2 x e dx 1 x is an answer, not a question. This is the solution– you don’t have to integrate it! The most important thing is setting up the integral! (Once you’ve done that, you can always evaluate it numerically on a computer.) So, why bother to evaluate integrals by hand, then? Because you often get families of related integrals, such as � ∞ x e F (a) = dx a x 1 where you want to ﬁnd how the answer depends on, say, a. 3 Lecture 30 18.01 Fall 2006 Arc Length This is very useful to know for 18.02 (multi-variable calculus). y ds y=f(x) dy dx x Figure 1: Inﬁnitesimal Arc Length ds ds dy dx Figure 2: Zoom in on Figure 1 to see an approximate right triangle. In Figures 1 and 2, s denotes arc length and ds = the inﬁnitesmal of arc length. � � 2 ds = (dx)2 + (dy)2 = 1 + (dy/dx) dx Integrating with respect to ds ﬁnds the length of a curve between two points (see Figure 3). To ﬁnd the length of the curve between P0 and P1 , evaluate: � P1 ds P0 4 Lecture 30 18.01 Fall 2006 P₁ P₀ a b Figure 3: Find length of curve between P0 and P1 . We want to integrate with respect to x, not s, so we do the same algebra as above to ﬁnd ds in terms of dx. � �2 (ds)2 (dx)2 (dy)2 dy = + =1+ (dx)2 (dx)2 (dx)2 dx Therefore, � P1 � b � ds = P0 � 1+ a dy dx �2 dx Example 5: The Circle. x2 + y 2 = 1 (see Figure 4). Figure 4: The circle in Example 1. 5 Lecture 30 18.01 Fall 2006 We want to ﬁnd the length of the arc in Figure 5: a Figure 5: Arc length to be evaluated. � 1 − x2 � � dy −2x 1 −x =√ =√ 2 dx 2 1−x 1 − x2 � � �2 −x ds = 1 + √ dx 1 − x2 � �2 −x x2 1 − x2 + x2 1 1+ √ =1+ = = 2 2 2 1−x 1−x 1 − x2 1−x � 1 ds = dx 1 − x2 � a ⏐a dx ⏐ √ s= = sin−1 x⏐ = sin−1 a − sin−1 0 = sin−1 a 2 0 1−x 0 sin s = a y= This is illustrated in Figure 6. 6 Lecture 30 18.01 Fall 2006 s a 1 a 1 Figure 6: s = angle in radians. Parametric Equations Example 6. x = a cos t y = a sin t Ask yourself: what’s constant? What’s varying? Here, t is variable and a is constant. Is there a relationship between x and y? Yes: x2 + y 2 = a2 cos2 t + a2 sin2 t = a2 Extra information (besides the circle): At t = 0, x = a cos 0 = a and y = a sin 0 = 0 π At t = , 2 π π x = a cos = 0 and y = a sin = a 2 2 Thus, for 0 ≤ t ≤ π/2, a quarter circle is traced counter-clockwise (Figure 7). 7 Lecture 30 18.01 Fall 2006 t=π/2 (0,a) (a,0) t=0 Figure 7: Example 6. x = a cos t, y = a sin t; the particle is moving counterclockwise. Example 7: The Ellipse See Figure 8. x = 2 sin t; y = cos t 2 x + y 2 = 1( =⇒ (2 sin t)2 /4 + (cos t)2 = sin2 t + cos2 t = 1) 4 t=0 (0,1) (2,0) t=π/2 Figure 8: Ellipse: x = 2 sin t, y = cos t (traced clockwise). Arclength ds for Example 6. dx = −a sin t dt, dy = a cos t dt � � � ds = (dx)2 + (dy)2 = (−a sin t dt)2 + (a cos t dt)2 = (a sin t)2 + (a cos t)2 dt = a dt 8