Use the substitution method to evaluate these

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Chapter 7 / Exercise 1
Calculus
Stewart
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use the substitution method to evaluate these integrands... Expression Change of Variable Identity to Use To Obtain p a 2 - x 2 x = a sin 1 - sin 2 = cos 2 a cos p a 2 + x 2 x = a tan 1 + tan 2 = sec 2 a sec p x 2 - a 2 x = a sec sec 2 - 1 = tan 2 a tan In order rewrite our answer in terms of the original variable x, we refer to the appropriate right angle triangle. Note: To use these substitutions, we have to make sure that is between a certain range so the function is injective . 30
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Calculus
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Chapter 7 / Exercise 1
Calculus
Stewart
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Janelle Resch’s MATH 138 Notes Spring 2014 Expression Change of Variable Interval p a 2 - x 2 x = a sin - 2 2 p a 2 + x 2 x = a tan - 2 < < 2 p x 2 - a 2 x = a sec 0 2 or 3 2 Let’s look at some examples... Example: Evaluate R 1 x 2 p x 2 +9 d x . Let’s use the trig substitution method. Let x = 3 tan d x = 3 sec 2 d Then we have x 2 + 9 = 9 tan 2 + 9 = 9(tan 2 + 1) = 9 sec 2 Making this change of variable gives Z 1 x 2 p x 2 + 9 d x = Z 1 (3 tan ) 2 p 3 2 sec 2 3 sec 2 d = 1 9 Z 3 sec 2 tan 2 3 sec d = 1 9 Z sec tan 2 d = 1 9 Z 1 cos sin 2 cos 2 d = 1 9 Z cos sin 2 d 31
Janelle Resch’s MATH 138 Notes Spring 2014 To evaluate this integral, we are going to have to use the substitution method again. u = sin d u = cos d = ) d u cos = d This gives 1 9 Z cos sin 2 d = 1 9 Z ⇠⇠ cos u 2 d u ⇠⇠ cos = 1 9 Z 1 u 2 d u = 1 9 - 1 u + C Putting this back in terms of we obtain 1 9 Z cos sin 2 d = - 1 9 sin + C = - 1 9 csc + C Now, we have to write our expression in terms of x. Recall that csc = hyp opp , thus csc = p 9 + x 2 x Therefore, we have that Z 1 x 2 p x 2 + 9 d x = - 1 9 p 9 + x 2 x + C. Example: Evaluate R x 2 (4 - x 2 ) 3 2 d x . Let’s use the trig substitution method. Let x = 2 sin d x = 2 cos d 32
Janelle Resch’s MATH 138 Notes Spring 2014 Then we have 4 - x 2 = 4 - 2 2 sin 2 = 4(1 - sin 2 ) = 4 cos 2 x Making this change of variable gives Z x 2 (4 - x 2 ) 3 2 d x = Z (2 sin ) 2 (2 2 cos 2 ) 3 2 2 cos d = Z 2 3 sin 2 cos (2 2 ) 3 2 (cos 2 ) 3 2 d = Z sin 2 ⇠⇠ cos cos 3 d = Z sin 2 cos 2 d = Z tan 2 d = Z (sec 2 - 1) d = tan - + C Now, we have to write our expression in terms of x. By our triangle, we see that tan = x p 4 - x 2 Finally, since - 2 2 , we have that = arcsin x 2 . Therefore, we have that Z x 2 (4 - x 2 ) 3 2 d x = x p 4 - x 2 - arcsin x 2 + C. 33
Janelle Resch’s MATH 138 Notes Spring 2014 End of Lecture 4 1. Evaluate the following integrals. State the method of integration you are using. (a) R 1 sec x tan 2 x d x (b) R x 3 p 9+ x 2 d x (b) R x 2 p 3 - x 2 d x 34
Janelle Resch’s MATH 138 Notes Spring 2014 MATH 138 - Lecture 5 Note: Sometimes, you will have to solve problems where the integrand has to be manipulated so it can be written in the form p a 2 - x 2 , p a 2 + x 2 or p x 2 - a 2 . For instance... Example: Evaluate R 1 x 2 - 6 x +25 d x . To solve such problems, we have to complete the square and then perform a change of variable. Consider, x 2 - 6 x + 25 = ( x - 3) 2 + 16 ( ) Since 16 is a perfect square, we know we can use trig substitution. Now, let’s make the following change of variable... y = ( x - 3) d y = d x Thus, we can rewrite (*) as ( x - 3) 2 + 16 = y 2 + 4 2 . Thus, Z 1 x 2 - 6 x + 25 d x = Z 1 y 2 + 16 d y Now consider y = 4 tan d y = 4 sec 2 d Then we have y 2 + 16 = 16 tan 2 + 16 = 16(tan 2 + 1) = 16 sec 2 35

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