Integration Techniques: Trigonometric Substitutions and Integration by Parts, Cheat Sheet of Mathematics

Download Integration Techniques: Trigonometric Substitutions and Integration by Parts and more Cheat Sheet Mathematics in PDF only on Docsity! 10 Techniques of Integration 10.1 Powers of sine and osine Functions consisting of products of the sine and cosine can be integrated by using substi- tution and trigonometric identities. These can sometimes be tedious, but the technique is straightforward. Some examples will suffice to explain the approach. EXAMPLE 10.1.1 Evaluate ∫ sin5 x dx. Rewrite the function: ∫ sin5 x dx = ∫ sinx sin4 x dx = ∫ sinx(sin2 x)2 dx = ∫ sinx(1− cos2 x)2 dx. Now use u = cosx, du = − sinx dx: ∫ sinx(1− cos2 x)2 dx = ∫ −(1− u2)2 du = ∫ −(1− 2u2 + u4) du = −u+ 2 3 u3 − 1 5 u5 + C = − cosx+ 2 3 cos3 x− 1 5 cos5 x+ C. 203 204 Chapter 10 Techniques of Integration EXAMPLE 10.1.2 Evaluate ∫ sin6 x dx. Use sin2 x = (1− cos(2x))/2 to rewrite the function: ∫ sin6 x dx = ∫ (sin2 x)3 dx = ∫ (1− cos 2x)3 8 dx = 1 8 ∫ 1− 3 cos 2x+ 3 cos2 2x− cos3 2x dx. Now we have four integrals to evaluate: ∫ 1 dx = x and ∫ −3 cos 2x dx = −3 2 sin 2x are easy. The cos3 2x integral is like the previous example: ∫ − cos3 2x dx = ∫ − cos 2x cos2 2x dx = ∫ − cos 2x(1− sin2 2x) dx = ∫ −1 2 (1− u2) du = −1 2 ( u− u3 3 ) = −1 2 ( sin 2x− sin3 2x 3 ) . And finally we use another trigonometric identity, cos2 x = (1 + cos(2x))/2: ∫ 3 cos2 2x dx = 3 ∫ 1 + cos 4x 2 dx = 3 2 ( x+ sin 4x 4 ) . So at long last we get ∫ sin6 x dx = x 8 − 3 16 sin 2x− 1 16 ( sin 2x− sin3 2x 3 ) + 3 16 ( x+ sin 4x 4 ) + C. EXAMPLE 10.1.3 Evaluate ∫ sin2 x cos2 x dx. Use the formulas sin2 x = (1−cos(2x))/2 and cos2 x = (1 + cos(2x))/2 to get: ∫ sin2 x cos2 x dx = ∫ 1− cos(2x) 2 · 1 + cos(2x) 2 dx. The remainder is left as an exercise. 10.2 Trigonometric Substitutions 207 First we do ∫ sec u du, which we will need to compute ∫ sec3 u du: ∫ secu du = ∫ secu secu+ tanu secu+ tanu du = ∫ sec2 u+ secu tanu sec u+ tanu du. Now let w = secu + tanu, dw = sec u tanu + sec2 u du, exactly the numerator of the function we are integrating. Thus ∫ secu du = ∫ sec2 u+ secu tanu sec u+ tanu du = ∫ 1 w dw = ln |w|+ C = ln | secu+ tanu|+ C. Now for ∫ sec3 u du: sec3 u = sec3 u 2 + sec3 u 2 = sec3 u 2 + (tan2 u+ 1) secu 2 = sec3 u 2 + secu tan2 u 2 + secu 2 = sec3 u+ secu tan2 u 2 + secu 2 . We already know how to integrate secu, so we just need the first quotient. This is “simply” a matter of recognizing the product rule in action: ∫ sec3 u+ sec u tan2 u du = secu tanu. So putting these together we get ∫ sec3 u du = sec u tanu 2 + ln | secu+ tanu| 2 + C, and reverting to the original variable x: ∫ √ 1 + x2 dx = secu tanu 2 + ln | secu+ tanu| 2 + C = sec(arctanx) tan(arctanx) 2 + ln | sec(arctanx) + tan(arctanx)| 2 + C = x √ 1 + x2 2 + ln | √ 1 + x2 + x| 2 + C, using tan(arctanx) = x and sec(arctanx) = √ 1 + tan2(arctanx) = √ 1 + x2. 208 Chapter 10 Techniques of Integration Exercises 10.2. Find the antiderivatives. 1. ∫ cscx dx ⇒ 2. ∫ csc3 xdx ⇒ 3. ∫ √ x2 − 1 dx ⇒ 4. ∫ √ 9 + 4x2 dx ⇒ 5. ∫ x √ 1− x2 dx ⇒ 6. ∫ x2 √ 1− x2 dx ⇒ 7. ∫ 1 √ 1 + x2 dx ⇒ 8. ∫ √ x2 + 2x dx ⇒ 9. ∫ 1 x2(1 + x2) dx ⇒ 10. ∫ x2 √ 4− x2 dx ⇒ 11. ∫ √ x √ 1− x dx ⇒ 12. ∫ x3 √ 4x2 − 1 dx ⇒ 10.3 Integration by Parts We have already seen that recognizing the product rule can be useful, when we noticed that ∫ sec3 u+ sec u tan2 u du = secu tanu. As with substitution, we do not have to rely on insight or cleverness to discover such antiderivatives; there is a technique that will often help to uncover the product rule. Start with the product rule: d dx f(x)g(x) = f ′(x)g(x) + f(x)g′(x). We can rewrite this as f(x)g(x) = ∫ f ′(x)g(x) dx+ ∫ f(x)g′(x) dx, and then ∫ f(x)g′(x) dx = f(x)g(x)− ∫ f ′(x)g(x) dx. This may not seem particularly useful at first glance, but it turns out that in many cases we have an integral of the form ∫ f(x)g′(x) dx but that ∫ f ′(x)g(x) dx is easier. This technique for turning one integral into another is called integration by parts, and is usually written in more compact form. If we let u = f(x) and v = g(x) then 10.3 Integration by Parts 209 du = f ′(x) dx and dv = g′(x) dx and ∫ u dv = uv − ∫ v du. To use this technique we need to identify likely candidates for u = f(x) and dv = g′(x) dx. EXAMPLE 10.3.1 Evaluate ∫ x lnx dx. Let u = lnx so du = 1/x dx. Then we must let dv = x dx so v = x2/2 and ∫ x lnx dx = x2 lnx 2 − ∫ x2 2 1 x dx = x2 lnx 2 − ∫ x 2 dx = x2 lnx 2 − x2 4 + C. EXAMPLE 10.3.2 Evaluate ∫ x sinx dx. Let u = x so du = dx. Then we must let dv = sinx dx so v = − cosx and ∫ x sinx dx = −x cosx− ∫ − cosx dx = −x cosx+ ∫ cosx dx = −x cosx+ sinx+ C. EXAMPLE 10.3.3 Evaluate ∫ sec3 x dx. Of course we already know the answer to this, but we needed to be clever to discover it. Here we’ll use the new technique to discover the antiderivative. Let u = sec x and dv = sec2 x dx. Then du = secx tanx dx and v = tanx and ∫ sec3 x dx = secx tanx− ∫ tan2 x secx dx = secx tanx− ∫ (sec2 x− 1) secx dx = secx tanx− ∫ sec3 x dx+ ∫ secx dx. 212 Chapter 10 Techniques of Integration 7. ∫ x arctan xdx ⇒ 8. ∫ x3 sinx dx ⇒ 9. ∫ x3 cosx dx ⇒ 10. ∫ x sin2 x dx ⇒ 11. ∫ x sin x cos x dx ⇒ 12. ∫ arctan( √ x) dx ⇒ 13. ∫ sin( √ x) dx ⇒ 14. ∫ sec2 x csc2 x dx ⇒ 10.4 Rational Fun tions A rational function is a fraction with polynomials in the numerator and denominator. For example, x3 x2 + x− 6 , 1 (x− 3)2 , x2 + 1 x2 − 1 , are all rational functions of x. There is a general technique called “partial fractions” that, in principle, allows us to integrate any rational function. The algebraic steps in the technique are rather cumbersome if the polynomial in the denominator has degree more than 2, and the technique requires that we factor the denominator, something that is not always possible. However, in practice one does not often run across rational functions with high degree polynomials in the denominator for which one has to find the antiderivative function. So we shall explain how to find the antiderivative of a rational function only when the denominator is a quadratic polynomial ax2 + bx+ c. We should mention a special type of rational function that we already know how to integrate: If the denominator has the form (ax + b)n, the substitution u = ax + b will always work. The denominator becomes un, and each x in the numerator is replaced by (u − b)/a, and dx = du/a. While it may be tedious to complete the integration if the numerator has high degree, it is merely a matter of algebra. 10.4 Rational Functions 213 EXAMPLE 10.4.1 Find ∫ x3 (3− 2x)5 dx. Using the substitution u = 3− 2x we get ∫ x3 (3− 2x)5 dx = 1 −2 ∫ ( u−3 −2 )3 u5 du = 1 16 ∫ u3 − 9u2 + 27u− 27 u5 du = 1 16 ∫ u−2 − 9u−3 + 27u−4 − 27u−5 du = 1 16 ( u−1 −1 − 9u−2 −2 + 27u−3 −3 − 27u−4 −4 ) + C = 1 16 ( (3− 2x)−1 −1 − 9(3− 2x)−2 −2 + 27(3− 2x)−3 −3 − 27(3− 2x)−4 −4 ) + C = − 1 16(3− 2x) + 9 32(3− 2x)2 − 9 16(3− 2x)3 + 27 64(3− 2x)4 + C We now proceed to the case in which the denominator is a quadratic polynomial. We can always factor out the coefficient of x2 and put it outside the integral, so we can assume that the denominator has the form x2 + bx+ c. There are three possible cases, depending on how the quadratic factors: either x2 + bx+ c = (x− r)(x− s), x2 + bx+ c = (x− r)2, or it doesn’t factor. We can use the quadratic formula to decide which of these we have, and to factor the quadratic if it is possible. EXAMPLE 10.4.2 Determine whether x2+x+1 factors, and factor it if possible. The quadratic formula tells us that x2 + x+ 1 = 0 when x = −1± √ 1− 4 2 . Since there is no square root of −3, this quadratic does not factor. EXAMPLE 10.4.3 Determine whether x2−x−1 factors, and factor it if possible. The quadratic formula tells us that x2 − x− 1 = 0 when x = 1± √ 1 + 4 2 = 1± √ 5 2 . Therefore x2 − x− 1 = ( x− 1 + √ 5 2 )( x− 1− √ 5 2 ) . 214 Chapter 10 Techniques of Integration If x2 + bx+ c = (x− r)2 then we have the special case we have already seen, that can be handled with a substitution. The other two cases require different approaches. If x2 + bx+ c = (x− r)(x− s), we have an integral of the form ∫ p(x) (x− r)(x− s) dx where p(x) is a polynomial. The first step is to make sure that p(x) has degree less than 2. EXAMPLE 10.4.4 Rewrite ∫ x3 (x− 2)(x+ 3) dx in terms of an integral with a numer- ator that has degree less than 2. To do this we use long division of polynomials to discover that x3 (x− 2)(x+ 3) = x3 x2 + x− 6 = x− 1 + 7x− 6 x2 + x− 6 = x− 1 + 7x− 6 (x− 2)(x+ 3) , so ∫ x3 (x− 2)(x+ 3) dx = ∫ x− 1 dx+ ∫ 7x− 6 (x− 2)(x+ 3) dx. The first integral is easy, so only the second requires some work. Now consider the following simple algebra of fractions: A x− r + B x− s = A(x− s) +B(x− r) (x− r)(x− s) = (A+B)x−As−Br (x− r)(x− s) . That is, adding two fractions with constant numerator and denominators (x−r) and (x−s) produces a fraction with denominator (x− r)(x− s) and a polynomial of degree less than 2 for the numerator. We want to reverse this process: starting with a single fraction, we want to write it as a sum of two simpler fractions. An example should make it clear how to proceed. EXAMPLE 10.4.5 Evaluate ∫ x3 (x− 2)(x+ 3) dx. We start by writing 7x− 6 (x− 2)(x+ 3) as the sum of two fractions. We want to end up with 7x− 6 (x− 2)(x+ 3) = A x− 2 + B x+ 3 . If we go ahead and add the fractions on the right hand side we get 7x− 6 (x− 2)(x+ 3) = (A+B)x+ 3A− 2B (x− 2)(x+ 3) . So all we need to do is find A and B so that 7x − 6 = (A + B)x+ 3A − 2B, which is to say, we need 7 = A+B and −6 = 3A− 2B. This is a problem you’ve seen before: solve a 10.5 Numerical Integration 217 the areas of all trapezoids we get f(x0) + f(x1) 2 ∆x+ f(x1) + f(x2) 2 ∆x+ · · ·+ f(xn−1) + f(xn) 2 ∆x = ( f(x0) 2 + f(x1) + f(x2) + · · ·+ f(xn−1) + f(xn) 2 ) ∆x. This is usually known as the Trapezoid Rule. For a modest number of subintervals this is not too difficult to do with a calculator; a computer can easily do many subintervals. xi xi+1 (xi, f(xi)) (xi+1, f(xi+1)) ............................................ ... ....... ............ .. Figure 10.5.2 A single trapezoid. In practice, an approximation is useful only if we know how accurate it is; for example, we might need a particular value accurate to three decimal places. When we compute a particular approximation to an integral, the error is the difference between the approxi- mation and the true value of the integral. For any approximation technique, we need an error estimate, a value that is guaranteed to be larger than the actual error. If A is an approximation and E is the associated error estimate, then we know that the true value of the integral is between A − E and A + E. In the case of our approximation of the integral, we want E = E(∆x) to be a function of ∆x that gets small rapidly as ∆x gets small. Fortunately, for many functions, there is such an error estimate associated with the trapezoid approximation. THEOREM 10.5.1 Suppose f has a second derivative f ′′ everywhere on the interval [a, b], and |f ′′(x)| ≤ M for all x in the interval. With ∆x = (b − a)/n, an error estimate for the trapezoid approximation is E(∆x) = b− a 12 M(∆x)2 = (b− a)3 12n2 M. Let’s see how we can use this. 218 Chapter 10 Techniques of Integration EXAMPLE 10.5.2 Approximate ∫ 1 0 e−x 2 dx to two decimal places. The second deriva- tive of f = e−x 2 is (4x2−2)e−x 2 , and it is not hard to see that on [0, 1], |(4x2−2)e−x 2 | ≤ 2. We begin by estimating the number of subintervals we are likely to need. To get two dec- imal places of accuracy, we will certainly need E(∆x) < 0.005 or 1 12 (2) 1 n2 < 0.005 1 6 (200) < n2 5.77 ≈ √ 100 3 < n With n = 6, the error estimate is thus 1/63 < 0.0047. We compute the trapezoid approxi- mation for six intervals: ( f(0) 2 + f(1/6) + f(2/6) + · · ·+ f(5/6) + f(1) 2 ) 1 6 ≈ 0.74512. So the true value of the integral is between 0.74512 − 0.0047 = 0.74042 and 0.74512 + 0.0047 = 0.74982. Unfortunately, the first rounds to 0.74 and the second rounds to 0.75, so we can’t be sure of the correct value in the second decimal place; we need to pick a larger n. As it turns out, we need to go to n = 12 to get two bounds that both round to the same value, which turns out to be 0.75. For comparison, using 12 rectangles to approximate the area gives 0.7727, which is considerably less accurate than the approximation using six trapezoids. In practice it generally pays to start by requiring better than the maximum possible error; for example, we might have initially required E(∆x) < 0.001, or 1 12 (2) 1 n2 < 0.001 1 6 (1000) < n2 12.91 ≈ √ 500 3 < n Had we immediately tried n = 13 this would have given us the desired answer. The trapezoid approximation works well, especially compared to rectangles, because the tops of the trapezoids form a reasonably good approximation to the curve when ∆x is fairly small. We can extend this idea: what if we try to approximate the curve more closely, 10.5 Numerical Integration 219 by using something other than a straight line? The obvious candidate is a parabola: if we can approximate a short piece of the curve with a parabola with equation y = ax2+bx+c, we can easily compute the area under the parabola. There are an infinite number of parabolas through any two given points, but only one through three given points. If we find a parabola through three consecutive points (xi, f(xi)), (xi+1, f(xi+1)), (xi+2, f(xi+2)) on the curve, it should be quite close to the curve over the whole interval [xi, xi+2], as in figure 10.5.3. If we divide the interval [a, b] into an even number of subintervals, we can then approximate the curve by a sequence of parabolas, each covering two of the subintervals. For this to be practical, we would like a simple formula for the area under one parabola, namely, the parabola through (xi, f(xi)), (xi+1, f(xi+1)), and (xi+2, f(xi+2)). That is, we should attempt to write down the parabola y = ax2 + bx + c through these points and then integrate it, and hope that the result is fairly simple. Although the algebra involved is messy, this turns out to be possible. The algebra is well within the capability of a good computer algebra system like Sage, so we will present the result without all of the algebra; you can see how to do it in this Sage worksheet. To find the parabola, we solve these three equations for a, b, and c: f(xi) = a(xi+1 −∆x)2 + b(xi+1 −∆x) + c f(xi+1) = a(xi+1) 2 + b(xi+1) + c f(xi+2) = a(xi+1 +∆x)2 + b(xi+1 +∆x) + c Not surprisingly, the solutions turn out to be quite messy. Nevertheless, Sage can easily compute and simplify the integral to get ∫ xi+1+∆x xi+1−∆x ax2 + bx+ c dx = ∆x 3 (f(xi) + 4f(xi+1) + f(xi+2)). Now the sum of the areas under all parabolas is ∆x 3 (f(x0)+4f(x1)+f(x2)+f(x2)+4f(x3)+f(x4)+ · · ·+f(xn−2)+4f(xn−1)+f(xn)) = ∆x 3 (f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) + · · ·+ 2f(xn−2) + 4f(xn−1) + f(xn)). This is just slightly more complicated than the formula for trapezoids; we need to remember the alternating 2 and 4 coefficients; note that n must be even for this to make sense. This approximation technique is referred to as Simpson’s Rule. As with the trapezoid method, this is useful only with an error estimate: