MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� Lecture 28 18.01 Fall 2006 Lecture 28: Integration by Inverse Substitution; Completing the Square Trigonometric Substitutions, continued -a0x aFigure 1: Find area of shaded portion of semicircle. � x � a2 − t2dt 0 t = a sin u; dt = a cos u du a 2 − t2 = a 2 − a 2 sin2 u = a 2 cos2 u = ⇒ a2 − t2 = a cos u (No more square root!) Start: x = −a ⇔ u = −π/2; Finish: x = a ⇔ u = π/2 � � � � �� a2 − t2 dt = a 2 cos2 u du = a 2 1 + cos(2u) du = a 2 u + sin(2u)+ c2 2 4 1 + cos(2u)(Recall, cos2 u = ).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). sin(2u) 2 sin u cos u 1 = = sin u cos u4 4 2 � � xsin 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). ax√a²-x²u2Figure 2: sin u = x/a; cos u = pa − x2/a. From the diagram, we see √a2 − x2 cos u = a And finally, � � �� � x � a2 − t2 dt = a 2 u + 1 sin u cos u − 0 = a 2 sin−1(x/a)+1 � x � √a2 − x2 4 2 2 2 a a0 x 2� a x 1 � a2 − t2 dt = 2 sin−1( a ) + 2 x a2 − x2 0 When the answer is this complicated, the route to getting there has to be rather complicated. There’s no way to avoid the complexity. 1Let’s double-check this answer. The area of the upper shaded sector in Figure 3 is a 2 u. The 2 area of the lower shaded region, which is a triangle of height √a2 − x2 and base x, is 1 x√a2 − x2.2 2� � � = � � � Lecture 28 18.01 Fall 2006 0xuFigure 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 substitution trig identity dx � √x2 + 1 x = tan u tan2 u + 1 = sec2 u dx √x2 x = sec u sec2 u − 1 = tan2 u � − 1 dx √1 − x2 x = sin u 1 − sin2 u = cos2 u Let’s extend this further. How can we evaluate an integral like this? dx √x2 + 4x When you have a linear and a quadratic term under the square root, complete the square. x 2 + 4x = (something)2 ± constant In this case, (x + 2)2 = x 2 + 4x + 4 = ⇒ x 2 + 4x = (x + 2)2 − 4 Now, we make a substitution. v = x + 2 and dv = dx Plugging these in gives us � � dx dv (x + 2)2 − 4 √v2 − 4 Now, let v = 2 sec u and dv = 2 sec u tan u dv 2 sec u tan u du √v2 = 2 tan u = sec u du − 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. 2 v = 2 sec u cos u = ⇔ v Set up a right triangle as in Figure 4. Express tan u in terms of v. v2√v²-4uFigure 4: sec u = v/2 or cos u = 2/v. Just from looking at the triangle, we can read off v √v2 − 4 sec u = and tan u = 2 2 2 sec u du = ln v + √v2 − 4+ c2 2 = ln(v + v2 − 4) − ln 2 + c We can combine those last two terms into another constant, c˜. dx � √x2 + 4x = ln(x + 2 + x2 + 4x) + ˜c 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 x − 1+ x + 2 dx = ln |x − 1| + 3 ln |x + 2| + c 4Lecture 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 x − 1 x + 2 (x − 1)(x + 2) x2 + 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.
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