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lec28 - Lecture 28 18.01 Fall 2006 Lecture 28 Integration...

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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 a 2 - t 2 dt 0 t = a sin u ; dt = a cos u du a 2 - t 2 = a 2 - a 2 sin 2 u = a 2 cos 2 u = a 2 - t 2 = a cos u (No more square root!) Start: x = - a u = - π / 2; Finish: x = a u = π / 2 a 2 - t 2 dt = a 2 cos 2 u du = a 2 1 + cos(2 u ) du = a 2 u + sin(2 u ) + c 2 2 4 1 + cos(2 u ) (Recall, cos 2 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(2 u ) 2 sin u cos u 1 = = sin u cos u 4 4 2 ( ) x sin u = sin sin - 1 ( x/a ) = a 1
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( ) 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 a ² -x ² u 2 Figure 2: sin u = x/a ; cos u = p a - x 2 /a . From the diagram, we see a 2 - x 2 cos u = a And finally, x a 2 - t 2 dt = a 2 u + 1 sin u cos u - 0 = a 2 sin - 1 ( x/a ) + 1 x a 2 - x 2 4 2 2 2 a a 0 x 2 a x 1 a 2 - t 2 dt = 2 sin - 1 ( a ) + 2 x a 2 - x 2 0 When the answer is this complicated, the route to getting there has to be rather complicated.
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