Calc1231W5T3 - Integration by circle trig substitution

Friday, 24 August 2018

1:05 PM

Untitled picture.png Machine generated alternative text: Let's evaluate the definite integral 10 1 1 dc x — 25 5 Note that we have to be a little careful here, as this is O on account of the denominator becoming a at one ot the end points an improper integral We use the circular trig substitution x = f(u) • the new lower bound a is the value such that f(a) • the new upper bound b is the value such that f(b) The integral can thus be re-written with respect to u: O _ As ranges from 5 to 107 the new variable u ranges from a to b, where: 5. This is a 10. This is b pi/3 1 b g(u) du where the integrand g(u) sec(u) O Hence the integral can be evaluated as 1 In (2+sqrt(3))              
Untitled picture.png Machine generated alternative text: Integrate using the substitution x 2 cos(u) O 1 The answer is 1 2 2 pi 2 dc Ink Drawings Ink Drawings Untitled picture.png Machine generated alternative text: Evaluate the definite integral using a suitable substitution. The answer is:                      Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings
Untitled picture.png Machine generated alternative text: Evaluate the definite integral using a suitable substitution. The answer is: Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings                             
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Machine generated alternative text: The rational function can be decomposed into partial fractions: Hence 10 (8x + 9) (8x — 5) 10 (8x + 9) (8x — 5) 96m + 10 = A(8x We can find A and B from this equation One way is to expand this equation and equate coefficients of x and x toget two linear equations. Anotherway is to observe that: • when x • when x 9 we see that A 8 5 we see that B 8 7 5 This partial fraction decomposition allows us to integrate the original function as 96X+10 1 J (8x+9) (8x — 5) 1 1 1 dC+B dc 0åt+C_
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 Machine generated alternative text: Let's integrate 1 We can use a substitution to express this integral in the form 1 The integral can then be evaluated as a function of u. But there are a few possibilities' Give your answers in terms of the variable u. • With the circle trig substitution x = 4sec(u) the integrand becomes dc —16 g(u) du. • With the hyperbolic trig substitution x 2 • With the substitution u • With the substitution u — 42 Hence I 4 cosh(u) the integrand becomes Hence I the integrand becomes Hence I — 42 the integrand becomes Hence I sqä(u) Any of these four answers can be re-expressed in terms of the original variable x, which leads to the conclusion: I = sqrt(xA2- 16)

 

 

 

 

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