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MSU MTH 132 - Mth132SupMat5-6

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Supplemental Material for Section 5.6.Simplifying Integrals by Substitutionby Richard O. Hill∗Introduction.Substitution is used throughout mathematics to simplify expressions so that they can be worked with moreeasily. Usually, we start by writing out all of the details of the substitution. Then, with proficiency, wewrite fewer and fewer details, perhaps for simple cases doing the whole substitution process in our heads.However, when work is to be graded, be sure to write down enough so that the grader can follow your work.We begin with one of the fundamental formulas of integration.(1)Zundu =un+1n + 1+ C, n 6= −1.This, of course, should be memorized.We give some examples.Example 1. FindZx2(2x3+ 5)2dx.Solution. Here we have a quantity to a power. We try to put this into the formZundu that we canapply Equation 1. The first principle of substitution is:(2) When you have a quantity to a power, let u equal that quantity and try to get du.Warning: u = the quantity, not the quantity to the power.Let u = 2x3+ 5. Then du = 6x2dxSince we are trying to get du, we first rewrite the problem with the x2next to the dx.Zx2(2x3+ 5)2dx =Z(2x3+ 5)2x2dx∗c°2002 Richard O. Hill. These notes are for the sole use of Michigan State University faculty and students. Any other userequires permission by the author.1Next multiply by66, bring the unwanted16outside the integral, and substitute in.Z(2x3+ 5)266x2dx =16Z(2x3+ 5)2| {z }u26x2dx| {z }du=16Zu2duWe can now use Formula 1 getting16Zu2du =16u33+ C =118(2x3+ 5)3+ CThis is the answer, which we can check by differentiating:·118(2x3+ 5)3+ C¸0=1183(2x3+ 5)2(6x2) + 0 = x2(2x3+ 5)2which is the original integrand, verifying that our answer is correct.Example 2. FindZ2 sin 3x[5 + cos 3x]4dx.Solution. Again we have a quantity to a power, so we let u equal the quantity and try to get du.u = 5 + cos 3x, du = −3 sin 3x dxWe work towards putting this in the formZundu. To avoid unnecessary confusion, we move the “2” outsidethe integral and rewrite this asZ2 sin 3x[5 + cos 3x]4dx = 2Z[5 + cos 3x]−4sin 3x dxWe proceed as before:= 2Z[5 + cos 3x]−4−3−3sin 3x dx= −23Z[5 + cos 3x]−4| {z }u−4(−3 sin 3x dx)| {z }du= −23Zu−4duWe use Formula 1 and get= −23u−3−3+ C =29[5 + cos 3x]−3+ C =29(5 + cos 3x)3+ CThis is the answer which you check by differentiating½29[5 + cos 3x]−3+ C¾0=29(−3)[5 + cos 3x]−4(−sin 3x · 3) + 0 =2 sin 3x[5 + cos 3x]−42Substitution with Trig Functions.There are four fundamental trig differentiation formulas:d(sin u) = cos u du d(tan u) = sec2u dud(cos u) = −sin u du d(sec u) = sec u tan u duBy going backwards, these lead to four fundamental integration formulas.(3)Zcos u du = sin u + CZsec2u du = tan u + CZsin u du = −cos u + CZsec u tan u du = sec u + CThere are similar formulas involving cot u and csc u.Example 3. FindZx3sin(5x4+ 6)dx.Solution. Here, we do not have a quantity to a power, so the first principle of substitution (2) does notapply. Hence, we turn to the second principle of substitution.(4) When you do not have a quantity to a power, but you do have a trig function of aquantity let u equal that quantity and try to get du.Later, this same principle will equally apply to other kinds of functions, such as exponential, logarithmic,inverse trig, etc.Let u = 5x4+ 6. Then du = 20x3dxWe work towards putting the problem in the formZsin u du. We first move the x3next to the dx.Zx3sin(5x4+ 6)dx =Zsin(5x4+ 6)x3dxNext, we multiply by2020, bring the unwanted120outside of the integral, and substitute inZsin(5x4+ 6)2020x3dx =120Zsin (5x4+ 6)| {z }u20x3dx| {z }du=120Zsin u du3We now use the appropriate formula from (2) getting120[−cos u] + C = −120cos(5x4+ 6) + CThis is the answer, which we can check by differentiating·−120cos(5x4+ 6) + C¸0= −120£−sin(5x4+ 6)(20x3)¤+ 0 = x3sin(5x4+ 6)Example 4. FindZ9 sec2(3√x + 5)√xdx.Solution. We recognize that we have a formula forZsec2u du in (3), so we move toward putting theproblem into this form.Let u = 3√x + 5 = 3x1/2+ 5. Then du =32x−1/2dx =321√xdxWe first move the “9” outside of the integral and move the1√xnext to the dx.Z9 sec2(3√x + 5)√xdx = 9Zsec2(3√x + 5)1√xdxNext multiply by23·32, move the unwanted23outside of the integral, simplify and substitute.9Zsec2(3√x + 5)23·321√xdx = 9 ·23Zsec2(3√x + 5)| {z }u32·1√xdx| {z }du= 6Zsec2u duWe apply the appropriate formula from (3) getting6 tan u + C = 6 tan(3√x + 5) + CThis is the answer, which you should check by differentiating, but we shall not here.Skipping Steps.The first five or so times you use substitution, you should write out all of the steps. Thereafter, you shouldskip steps that you can easily do in your head and that leaving out does not lead to errors.For instance, in Example 1,Zx2(2x3+ 5)2dx, we let u = 2x3+ 5 and find du = 6x2dx. Many peoplewould not physically move the x2next to the dx, nor would they multiply by66, but they would just thinkthese steps. They would proceed by:Zx2(2x3+ 5)2dx =16Z6x2(2x3+ 5)2dx =16Zu2du =16u33+ C =118(2x3+ 5)3+ C4When you are really experienced and the problem is simple enough, you might want to try doing morein your head. But be careful, and, again, if the work is being graded, make sure you write down enoughso that the grader can follow what you do. The balance is, as it is throughout mathematics, you want towrite down enough details so that you do not make errors, but not too many details so that the flow of theproblem is disrupted in your mind.Special Case.Very simple problems likeZcos 7x dx should be handled in a simple way. Remember, you knowZx7dx =18x8+ C and not x8+ C. You need the18to cancel the “8” when checking (x8)0= 8x7. This is a similarsituation. You knowZcos x dx = sin x +C. As withZx7dx,Zcos 7x dx =17sin 7x+C and not sin 7x +C,because when you check, (sin 7x)0= 7 cos 7x and you need the17to cancel the 7. Similarly,Zsecx2tanx2dx = 2 secx2+ Csince the “2” is needed to cancel the “12” when you check the answer.When Substitution Does Not Work.Sometimes the “obvious” choice of substitution does not work. This method of substitution allows you tosee as soon as possible that it does not work, and you must look for an alternative method.Example 5. FindZx(2x3+ 5)2dx (instead ofZx2(2x3+ 5)2dx of Example 1).Solution. We see we have a quantity to a power, so (as in Example 1) we try,u = 2x3+ 5, du = 6x2dx.But right here we see we cannot get x2since we only have an x. (Warning: We cannot multiply


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