Integration by Inverse Substitution; Completing the Square

Lecture 28 18.01 Fall 2006 Lecture 28: Integration by Inverse Substitution; Completing the Square Trigonometric Substitutions, continued -a 0 x a Figure 1: Find area of shaded portion of semicircle. � x � a2 − t2 dt 0 t = a sin u; dt = a cos u du � a2 − t2 = a2 − a2 sin2 u = a2 cos2 u =⇒ a2 − t2 = a cos u (No more square root!) Start: x = −a ⇔ u = −π/2; Finish: x = a ⇔ u = π/2 � � � � � � 1 + cos(2u) u sin(2u) a2 − t2 dt = a2 cos2 u du = a2 du = a2 + +c 2 2 4 1 + cos(2u) ). 2 We want to express this in terms of x, not u. When t = 0, a sin u = 0, and therefore u = 0. When t = x, a sin u = x, and therefore u = sin−1 (x/a). (Recall, cos2 u = sin(2u) 2 sin u cos u 1 = = sin u cos u 4 4 2 � � x sin u = sin sin−1 (x/a) = a 1 Lecture 28 18.01 Fall 2006 � � How can we find cos u = cos sin−1 (x/a) ? Answer: use a right triangle (Figure 2). a x u √a²-x² Figure 2: sin u = x/a; cos u = From the diagram, we see cos u = p a2 − x2 /a. √ a 2 − x2 a And finally, � x � a2 − t2 2 � dt = a 0 � 0 x � � √ � −1 u 1 (x/a) 1 � x � a2 − x2 2 sin + sin u cos u − 0 = a + 4 2 2 2 a a � a2 x 1 � a2 − t2 dt = sin−1 ( ) + x a2 − x2 2 a 2 When the answer is this complicated, the route to getting there has to be rather complicated. There’s no way to avoid the complexity. 1 Let’s double-check this answer. The area of the upper shaded sector in Figure 3 is a2 u. The 2 √ √ area of the lower shaded region, which is a triangle of height a2 − x2 and base x, is 12 x a2 − x2 . 2 Lecture 28 18.01 Fall 2006 u x 0 Figure 3: Area divided into a sector and a triangle. Here is a list of integrals that can be computed using a trig substitution and a trig identity. integral � dx √ x2 + 1 � dx √ x2 − 1 � dx √ 1 − x2 substitution trig identity x = tan u tan2 u + 1 = sec2 u x = sec u sec2 u − 1 = tan2 u x = sin u 1 − sin2 u = cos2 u Let’s extend this further. How can we evaluate an integral like this? � dx √ 2 x + 4x When you have a linear and a quadratic term under the square root, complete the square. x2 + 4x = (something)2 ± constant In this case, (x + 2)2 = x2 + 4x + 4 =⇒ x2 + 4x = (x + 2)2 − 4 Now, we make a substitution. v =x+2 and dv = dx Plugging these in gives us � dx � = (x + 2)2 − 4 � dv √ v2 − 4 Now, let � v = 2 sec u and dv = 2 sec u tan u � � dv 2 sec u tan u du √ = = sec u du 2 tan u v2 − 4 3 Lecture 28 18.01 Fall 2006 Remember that � sec u du = ln(sec u + tan u) + c Finally, rewrite everything in terms of x. v = 2 sec u ⇔ cos u = 2 v Set up a right triangle as in Figure 4. Express tan u in terms of v. v √v²-4 u 2 Figure 4: sec u = v/2 or cos u = 2/v. Just from looking at the triangle, we can read off √ v2 − 4 and tan u = 2 � � √ � v v2 − 4 2 sec u du = ln + +c 2 2 � = ln(v + v 2 − 4) − ln 2 + c v sec u = 2 We can combine those last two terms into another constant, c̃. � � dx √ = ln(x + 2 + x2 + 4x) + c̃ x2 + 4x Here’s a teaser for next time. In the next lecture, we’ll integrate all rational functions. By “rational functions,” we mean functions that are the ratios of polynomials: P (x) Q(x) It’s easy to evaluate an expression like this: � � � 1 3 + dx = ln |x − 1| + 3 ln |x + 2| + c x−1 x+2 4 Lecture 28 18.01 Fall 2006 If we write it a bit differently, however, it becomes much harder to integrate: 1 3 x + 2 + 3(x − 1) 4x − 1 + = = 2 x−1 x+2 (x − 1)(x + 2) x +x−2 � 4x − 1 = ??? x2 + x − 2 How can we reorganize what to do starting from (4x − 1)/(x2 + x − 2)? Next time, we’ll see how. It involves some algebra. 5