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18.01 Calculus Jason Starr Fall 2005 Lecture 26. November 18, 2005 Homework. Problem Set 7 Part I: (f)–(g); Part II: Problem 1 and Problem 2 (a), (b). Practice Problems. Course Reader: 5E-8, 5E-10, and please read through Part II of Problem Set 7.� � � � � 18.01 Calculus Jason Starr Fall 2005 1. Review of inverse substitution and another example. Recall the general strategy for finding an antiderivative of the form, F (x, √Ax2 + Bx + C) dx. G(x, √Ax2 + Bx + C) For definiteness, consider the example, 2xdx,√x2 − 2ax + 2a2 where a is a constant. Step 1. Complete the square. Complete the square of the expression Ax2 + Bx + C, inside the radical. In the example, 2 2 x − 2ax + 2a 2 = (x − a)2 + a . Step 2. Make a linear change of coordinates. Make a linear change of coordinates to simplify 2 2 2 2the quadratic term to one of the 3 types: a2 − x , x − a , or x2 + a . In the example, this means making the linear change of variables, u = x − a, du = dx. 2The new quadratic term is u2 + a , the third type. The new antiderivative is, 2(u + a)2 u2 + 2u + adu = du. 22√u2 + a√u2 + aStep 3. Use inverse substitution to e liminate the radicals. There is a choice of inverse sub-stitution: trigonometric, hyperbolic or rational. When starting out, it is a good idea to experiment with all 3. On an exam, usually one choice will be suggested (or even demanded). When no other guidance is given, trigonometric substitution is a good starting point (because you are already very familiar with trigonometric functions). In the example, to eliminate the radical, the correct inverse trigonometric substitution is, u = a tan(θ), du = a sec 2(θ)dθ. This is because the quadratic term becomes, 2 2 2 2 2 2u + a = a tan (θ) + a 2 = a sec (θ). With this substitution, the new antiderivative is, 2a2 tan2(θ) + 2a2 tan(θ) + a� a sec 2(θ)dθ. a2 sec2(θ)� � � � � � � � � � � � 18.01 Calculus Jason Starr Fall 2005 This simplifies to, 2 a (tan 2(θ) + 2 tan(θ) + 1) sec(θ)dθ. This can be written as a sum of 3 terms, 2 2a tan (θ) sec(θ)dθ + 2a 2 sec(θ) tan(θ)dθ + a 2 sec(θ)dθ. Step 4. Compute the new antiderivative. If this were only as simple as it sounds, how much easier calculus would be! This step is often difficult in itself. Often it requires at least one more direct substitution. Sometimes, it also requires a partial fractions decomposition. We will return to this step below. Step 5. Back-substitute. This is always a step for a method using direct substitution or inverse substitution. This step frequently introduces terms like cos(tan−1(x)). Time-p ermitting (or when specifically instructed to do so), these terms should be simplifed using the right-triangle method from lecture, 2θ = tan−1(x), x/1 = tan(θ) = Opposite/Adjacent, Hypotenuse = √1 + x , cos(θ) = Adjacent/Hypotenuse = Step 6. Check your answer. When feasible, check your answer. Since differentiation is so much faster than antidifferentiation, it is usually quite easy to check an antiderivative is correct. 1/√x2 .1 + Example. The tricky part is, of course, Step 4. In the example, the integral broke into 3 terms, 2 2a tan (θ) sec(θ)dθ + 2a 2 sec(θ) tan(θ)dθ + a 2 sec(θ)dθ. The last antiderivative was actually Problem 3(b) from Part II of Problem Set 4. It turns out to be, 2 a sec(θ)dθ = a 2 ln(u + √u2 + a2) + C = a2 ln(x − a + √x2 − 2ax a2C.+ 2 ) + The middle antiderivative is simply the derivative of sec(θ) = 1 + tan2(θ). So the middle term is, 2a 2 sec(θ) tan(θ)dθ = 2a 2 sec(θ) + C = 2a√a2 + u2 + C = 2a√x2 − 2ax a2 + C.+ 2But the final term does not simplify in an obvious way. In such cases, it is best to express everything in terms of sin(θ) and cos(θ) to get a fresh perspective, 2 2a tan (θ) sec(θ)dθ = a 2 sin2(θ) dθ. cos3(θ)� � � � � � 18.01 Calculus Jason Starr Fall 2005 Multiplying numerator and denominator by cos(θ) and expressing in terms of sin(θ) gives, sin2(θ)2 (cos2(θ))2 cos(θ)dθ = a 2 sin2(θ) a (1 − sin2(θ))2 cos(θ)dθ. Now substitute f or sin(θ), z = sin(θ), dz = cos(θ)dθ. The new antiderivative is, � 2zdz. (1 − z2)2 How do we compute this antiderivative? That is the topic of partial fractions. Remark: In lecture the solution was done a bit differently. This led to a slightly different an-tiderivative, 1 dz. (1 − z2)2 Notice the difference of these 2 antiderivatives is, (1) − (z2) 1 dz = dz. (1 − z2)2 (1 − z2) This was computed in Problem 3(a), Part II of Problem Set 4. Thus, computing either of the 2 antiderivatives gives both of them. 2. Antidifferentiating simple rational expressions. A rational expression is a fraction of polynomials, F (x)/G(x). These frequently arise in Step 4 of the algorithm above. From the point of view of antidifferentiation, the simplest rational expressions are either polynomials, q(x) = anx n + an−1x n−1 + a1x + a0,+ ···or else partial fractions, A . (x − a)m There are 2 other kinds of partial fractions which were not emphasized in lecture, B(x − a) C and . ((x − a)2 + b2)m ((x − a)2 + b2)m These 2 kinds come up less often than the first kind. But they do come up, for instance, when studying Laplace transforms in 18.03. Both polynomials and partial fractions are (relatively) easy to antidifferentiate. The antiderivative of a polynomial is, q(x)dx = an (nxn+1 + an−1 n xn + + a1 2 x2 + a0x + C.+1) ···� � � � � � 18.01 Calculus Jason Starr Fall 2005 The antiderivative of the first kind of partial fraction is, (−A/(m − 1))(x −a)−(m−1) + C , m ≥ 2,A x − a)−mdx = ( A ln( x − a ) + C , m = 1 | |The second kind of partial fraction can be computed with a direct substitution v = (x − a)2 + b2 , dv = 2(x − a)dx, B(x − a) B dv (− (2m − x − a)2 + b2)−(m−1) + C , ( x − a)2 + b2C , B/ 2))((B/2) ln(( ) + m ≥ 2, dx = = ((x − a)2 + b2)m 2 vm m = 1 The third kind of partial fraction can be computed with an inverse substitution x = b tan(θ) + a, dx = b sec2(θ)dθ, C dx = ((x − a)2 + b2)m Integration by parts gives a reduction formula for such integrals; see Problems (i) and (j), Part I of Problem Set 7. (2m−1) � cos2m−2(θ)dθ.C/b3. Simplifying rational expr essions: division and factoring Many rational expressions that come up are not of the simple kinds above. The


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