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 
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Untitled picture.png Machine generated alternative text:
Evaluate the definite integral 
using a suitable substitution. The answer is: 





















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Untitled picture.png Machine generated alternative text:
Evaluate the definite integral 
using a suitable substitution. The answer is: 
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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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