Lecture5 - Lecture 5 MATH 138-W12-003 I Trigonometric substitution Section 7.3 √ √ In this lecture we focus on computing integrals involving

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Unformatted text preview: Lecture 5: MATH 138-W12-003 January 13, 2012 I) Trigonometric substitution, Section 7.3 √ √ In this lecture we focus on computing integrals involving functions that include a2 − x2 , x2 − a2 √ and x2 + a2 using trigonometric substitutions. For example the first occurs when we want to compute the area of an ellipse. Example If the equation of an ellipse is x2 /a2 + y 2 /b2 = 1 then we can write the are as, a A=4 0 b√ 2 a − x2 dx. a At the moment we do not have the techniques for evaluating this. The new technique that we begin to develop is to make a trigonometric substitution of the form, x = a sin θ and dx = a cos θdθ. The textbook calls this an inverse substitution because we are making this more complicated but it is still a substitution so that’s the terminology I’m going to use. There is a geometric interpretation of this substitution. We are defining a triangle that has one √ side x, another a2 − x2 and a hypotenuse of a. We can assume that the angle we are dealing with is in the range −π/2 ≤ θ ≤ π/2. If x is positive than so is the angle but if x is negative, then the angle is negative. It is important to recall that cos θ is not negative in this interval so that | cos θ| = cos θ. To evaluate the integral we substitute our change of variables into the integral to yield after using our trigonometric substitution, a b a A=4 0 a2 − a2 sin2 θ a cos θdθ, π /2 1 − sin2 θ cos θdθ, = 4ab 0 sub in trig identity π /2 | cos θ| cos θdθ, = 4ab 0 but cos is positive π /2 cos2 θdθ. = 4ab 0 This integral is something we know how to solve by using a double angle substitution, π /2 A = 2ab 0 1 (1 + cos 2θ)dθ = 2ab θ + sin 2θ 2 π /2 = πab 0 Note that in the special case where a = b then we recover the formula for the area of a circle. The strategy that we should use is the following depending on the form of the square root. √ 1) If we have a2 − x2 then we make the substitution, x = a sin θ, − π π ≤θ≤ , 2 2 and we need to use the identity 1 − sin2 θ = cos2 θ. The square root always takes the form √ a2 − x2 = a| cos θ| = a cos θ. 1 2) If we have √ a2 + x2 then we make the substitution, − x = a tan θ, π π ≤θ≤ , 2 2 and we need to use the identity 1 + tan2 θ = sec2 θ. The square root always takes the form √ a2 + x2 = a| sec θ| = a sec θ. √ 3) If we have x2 − a2 then we make the substitution, x = a sec θ, 0≤θ≤ π 2 or π≤θ≤ 3π , 2 and we need to use the identity sec2 θ − 1 = tan2 θ. In the above domain, the first is for x √ positive and the second for x negative. The square root always takes the form x2 − a2 = a| tan θ| = a tan θ. a ￿ x x sin θ = , a x sin θ = √ cos θ = √ a2 − x2 , a tan θ = √ θ ￿ a2 − x2 a2 + x2 θ a2 x , + x2 cos θ = √ x2 − a2 , x a cos θ = , x a2 a , + x2 x a2 − x2 tan θ = x a a ￿ x θ x2 − a2 sin θ = √ tan θ = √ x2 − a2 x a Figure 1: In this Figure we see the three different types of triangles that we have depending on which case we need to consider. Example: Find √xx+4 dx. 2 Before we jump into using a trigonometric substitution we should stop and think about the integrand. In the denominator we have the square root of a quadratic. The derivative of a quadratic is proportional to x, which is what appears in the numerator. Therefore, there are two methods at our disposal. 2 Method 1 The simplest option is to substitute u = x2 + 4, du = 2xdx, which yields, √ √ x 1 du √ √ = u + C = x2 + 4 + C. dx = 2 u x2 + 4 Method 2 A trigonometric substitution that√ works for this problem is x = 2 tan θ and dx = 2 2 sec θdθ. Note that this implies that cos θ = 2/ x2 + 4. From this we can evaluate the integral in question, 2 tan θ √ 2 sec2 θdθ, 2 4 tan θ + 4 tan θ =2 √ 2 sec2 θdθ, now use trig identity tan θ + 1 tan θ =2 sec2 θdθ, but cos is positive | sec θ| x √ dx = x2 + 4 =2 tan θ sec θdθ, this is a standard integral = 2 sec θ + C, substitute in for θ, √ = x2 + 4 + C. Of course the two methods yield the same answer, as they must. Example: Evaluate √xdx a2 for a > 0. We begin with the substitution, x = a sec θ, dx = 2− a sec θ tan θdθ on the intervals 0 < θ < π/2 or π < θ < 3π/2. In either of these intervals the square root can be reduced as, √ √ √ √ x2 − a2 = a2 sec2 θ − a2 = a sec2 θ − 1 = a tan2 θ = a| tan θ| = a tan θ. We can substitute this into the integral to obtain, dx √ = x 2 − a2 = a sec θ tan θ dθ, a tan θ sec θdθ, = ln | sec θ + tan θ| + C. The last lines arises from a formula that we derived in Lecture 3. Given the triangle we can now rewrite this in terms of x, √ √ x x2 − a2 dx √ = ln | + | + C = ln |x + x2 − a2 | + C. a a x 2 − a2 In the above we used the property ln(ab) = ln a + ln b and absorbed − ln a into √ constant. the Question: What do we do if we have an integrand that has a function like ax2 + bx + c? Solution: Compute the square, √ ax2 + bx + c = = b a(x2 + x) + c, a b b2 a(x2 + )2 + c − . 2a 4a Then depending on the sign of c − b2 /(4a) we can use one of our previously mentioned methods. 3 ...
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This note was uploaded on 02/14/2012 for the course MATH 138 taught by Professor Anoymous during the Fall '07 term at Waterloo.

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