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# lec28 - MIT OpenCourseWare http/ocw.mit.edu 18.01 Single...

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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 .

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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 udu 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 ± ± ± ² ³ a 2 t 2 dt = a 2 cos 2 udu = 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 ´ µ x sin u = sin sin 1 ( x/a ) = a 1
± ² Lecture 28 18.01 Fall 2006

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lec28 - MIT OpenCourseWare http/ocw.mit.edu 18.01 Single...

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