Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://www.math.iastate.edu/hentzel/class.166.ICN Textbook: Calculus Early Vectors Preliminary Edition By Stewart. Monday, Feb 3 6.5 Substitution p407: (54),56,57,60,63,66,67,(68),73,75,77,(78),81,82,85, Main Idea: Don't forget the tail. Key Words: Even Functions, Odd Functions. Goal: Learn to integrate by exploiting the chain rule. Previous Assignment: p407, (20),(32),(46) p407 Problem 20 x^2 INT ---------- dx Sqrt[1-x] u = 1-x du = -dx (1-u)^2 INT ------------(-du) u^(1/2) -1 + 2 u - u^2 INT --------------- du u^(1/2) INT -u^(-1/2) + 2 u^(1/2) - u^(3/2) du -u^(1/2) u^(3/2) -u^(5/2) ----------- + 2 ---------- + ---------- + c 1/2 3/2 5/2 -2 u^(1/2) + 4/3 u^(3/2) -2/5 u^(5/2) + c Sqrt[1-x](-2 + 4/3 (1-x) -2/5 (1-x)^2 ) + c <====== ans Check: -2 (1-x)^(1/2) + 4/3 (1-x)^(3/2) -2/5 (1-x)^(5/2) + c - (1-x)^(-1/2)(-1) + 2(1-x)^(1/2)(-1) -(1-x)^(3/2)(-1) (1-x)^(-1/2) ( 1 - 2(1-x) + (1-x)^2 ) 1 - 2+2x +1-2x+x^2 ---------------------- Sqrt[1-x] x^2 -------------- Sqrt[1-x] ----------------------------------------------------------------- p407 Problem 32 INT (x^3+1)^(1/3) x^5 dx u = x^3 + 1 du = 3x^2 dx INT u^(1/3) x^5 du/(3x^2) du (1/3) INT u^(1/3) x^3 du (1/3) INT u^(1/3) (u-1) du (1/3) INT u^(4/3) - u^(1/3) du u^(7/3) u^(4/3) (1/3) (------------ - ---------- ) + c 7/3 4/3 (1/7) u^(7/3) - (1/4) u^(4/3) + c (1/7) (x^3+1)^(7/3) - 1/4 (x^3 + 1)^(4/3) + c <======== ans check: 1/3 (x^3+1)^(4/3)(3x^2) -1/3 (x^3+1)^(1/3) 3x^2 1/3 (x^3+1)^(1/3) [ (x^3+1)3x^2 - 3x^2 ] 1/3 (x^3+1)^(1/3) (x^3) 3x^2 (x^3+1)^(1/3) x^5. ------------------------------------------------------------------ p407 Problem 46 Sin[x] INT --------------- dx 1+Cos[x]^2 u = Cos[x] du = -Sin[x] dx -du INT ------------------ 1+ u^2 - ArcTan[u] + c -ArcTan[Cos[x]] + c <========== ans -Sin[x] Check - ----------------- 1+ Cos[x]^2 ------------------------------------------------------------------ New Material: Sometimes one has to poke around in the dark trying one thing after another in order to find the key to integration. Most of the time it is so obvious what the substitution should be, that one does not even have to go through the formal process of substitution. -------------------------------------------------------------- p407 Problem 28 ax+b INT ---------------------- dx Sqrt[ax^2 + 2 bx + c] We notice that we can make the numerator du by multiplying by 2 2ax+2b (1/2) INT ---------------------- dx Sqrt[ax^2 + 2 bx + c] 1/2 (a x^2+2 b x + c) 1/2 1/2 ----------------- = (a x^2+2 b x + c) + c 1/2 -------------------------------------------------------------- p407 Problem 48 x INT -------- dx 1+x^4 1 Remember, the derivative of ArcTan[u] = ----- 1+u^2 x INT -------- dx 2 1+(x^2) Adjust the numerator to made it the derivative of x^2, i.e. 2x dx 2x 1/2 INT -------- dx = 1/2 ArcTan[x^2] + C 2 1+(x^2) -------------------------------------------------------------- Some nice things to know about calculating some integrals. a INT anything = 0. a 3 INT Sqrt[1+x^4] dx = 0 3 A function is called "odd" if f(-x) = -f(x). Such functions are "reflected through the origin so that the curve to the left of zero is the same as the curve to the right of zero, except it is upside down. Examples are y = Sin[x] and y = x^3. In fact y = x^n where n is odd is also an example. If one takes a equal amount to the left and to the right of zero, the areas will cancel. a INT f(x) dx = 0 if y = f(x) is an odd function. -a ------------------------------------------------------------- a x=a INT Sin[x] dx = -Cos[x] | = - Cos[a] -(-Cos[-a]) = 0 -a x=-a x=a a x^4 | a^4 (-a)^4 INT x^3 dx = ----- | = ----- - -------- = 0. -a 4 x=-a 4 4 ----------------------------------------------------------------- A function is called "even" if f(-x) = +f(x). Such functions are "reflected through the y-axis so that the curve to the left of zero is the same as the curve to the right of zero, except it is backwards. Examples are y = Cos[x] and y = x^4. In fact y = x^n where n is even is also an example. Since the curves are identical as they depart from zero, the area from -a to zero is the same as the area from zero to a. --------------------------------------------------------------- 0 x=0 INT Cos[t] dt = Sin[t] | = Sin[0]-Sin[-a] = Sin[a] -a x=-a a x=a INT Cos[t] dt = Sin[t] | = Sin[a]-Sin[0] = Sin[a] 0 x=0 ---------------------------------------------------------- When you are using the definite integrals, there are two ways to handle the limits. You can change the x-endpoints to the u-endpoints when you change the x in to u's, OR you can compute the antiderivative y = F(x) and substitute in the x-endpoints. Be careful to avoid using the x-endpoints where the u-endpoints are required, and vis-versa. Page 405 Example 7: 4 INT Sqrt[3x+4] dx 0 x=4 INT Sqrt[3x+4] dx x=0 u = 3x+4 du = 3 dx 3/2 |u=16 u=16 u | INT u^(1/2) du/3 = 1/3 ----- | = (2/9)(64-8) = 112/9 u=4 3/2 | u=4 ----------------------------------------------------------------- Page 408 Problem 82: x=4 x=2 If f is continuous and INT f(x) dx = 10, find INT f(2x) dx. x=0 x=0 Let u = x/2, then du = 1/2 dx x=4 u=2 2 10 = INT f(x) dx = INT f(2u) 2 du = 2 INT f(2u) du x=0 u=0 u=0 x=2 So INT f(2x) dx = 5. x=0 ------------------------------------------------------------ Page 407 Problem 64 x=+Pi/2 x^2 Sin[x] INT --------------- dx x=-Pi/2 1+x^6 ----------------------------------------------------------------- Plot[x^2 Sin[x]/(1+x^6),{x,-Pi/2,Pi/2}] We shall use Mathematica to integrate the function. It will be too complicated to look at algebraically, but we can plot the integral to see what it looks like. In particular, we can see if the integral is zero like it should be. F[x_] := Integrate[x^2 Sin[x]/(1+x^6),x] F[x_] := (0.0562660772038117 - 0.1431633890500229*I)* CosIntegral[-0.8660254037844387 - 0.4999999999999999*I + x] + (0.05626607720381167 + 0.143163389050023*I)* CosIntegral[-0.8660254037844385 + 0.5000000000000005*I + x] - (0.1958668656073003 + 9.785515678127366*^-17*I)*CosIntegral[-1.*I + x] - (0.1958668656073003 + 9.785515678127366*^-17*I)*CosIntegral[1.*I + x] + (0.05626607720381164 + 0.143163389050023*I)* CosIntegral[0.8660254037844385 - 0.5000000000000005*I + x] + (0.05626607720381172 - 0.1431633890500229*I)* CosIntegral[0.8660254037844387 + 0.4999999999999999*I + x] - (0.06615825837150533 + 0.1217571698428708*I)* SinIntegral[-0.8660254037844387 - 0.4999999999999999*I + x] - (0.06615825837150557 - 0.1217571698428709*I)* SinIntegral[-0.8660254037844385 + 0.5000000000000005*I + x] - (1.284872737218814*^-16 - 0.2571801058025407*I)*SinIntegral[-1.*I + x] + (1.284872737218814*^-16 - 0.2571801058025407*I)*SinIntegral[1.*I + x] + (0.06615825837150558 - 0.1217571698428709*I)* SinIntegral[0.8660254037844385 - 0.5000000000000005*I + x] + (0.06615825837150536 + 0.1217571698428708*I)* SinIntegral[0.8660254037844387 + 0.4999999999999999*I + x] You notice that the graph is not zero at x = Pi/2. This is because the graph is not zero at x = -Pi/2. We want the area function for the curve that is zero at the start. We can get that area function by using F[x]-F[-Pi/2]. This area function gives the correct value of zero at -Pi/2 and Pi/2. The area becomes less than zero and then as the curve passes through the origin, it starts gaining back area and finally reaches zero area at Pi/2.