Substituting Integration with trigonometric function

  • #1

Homework Statement



[itex]\int_{1}^{5}\sqrt{4-(x-3)^2}dx[/itex]

Homework Equations





The Attempt at a Solution


Not really a homework. Just curious.

Let [itex]x-3=2sin u[/itex]
[itex]x=2sin u+3[/itex]

Now limit change,
when x=1, u = arcsin (-1)
when x=5, u = arcsin (1)

Since there are infinitely many solution of arcsin (-1) and arcsin (1), How do we know which ones to choose?

Thank you very much
 
Last edited:

Answers and Replies

  • #2
94
1
[itex]\int_{1}^{5}\sqrt{4-(x-3)^2}dx[/itex]

You have to write itex for the Latex to work, not tex. And the ending tag is /itex with a forward slash, not \itex with a backward slash.

Also, show your attempt at a solution or nobody will help you.
 
  • #3
94
1
There are not infinitely many solutions to [itex]sin^{-1}(-1)[/itex] and [itex]sin^{-1}(1)[/itex]. The range of [itex]sin^{-1}(x)[/itex] is [itex][-\frac{π}{2},\frac{π}{2}][/itex].

EDIT: While trigonometric substitution is a valid way to solve this definite integral, you can also solve the problem by thinking about what the graph of [itex]y=\sqrt{4-(x-3)^{2}}[/itex] is, and remembering that a definite integral is the area between the curve and the x-axis.
 
Last edited:
  • #4
There are not infinitely many solutions to [itex]sin^{-1}(-1)[/itex] and [itex]sin^{-1}(1)[/itex]. The range of [itex]sin^{-1}(x)[/itex] is [itex][-\frac{π}{2},\frac{π}{2}][/itex].

Why is that? can't it be [3∏/2, 5∏/2]? or let's say [-∏/2 , 5∏/2] ?
 
  • #5
There are not infinitely many solutions to [itex]sin^{-1}(-1)[/itex] and [itex]sin^{-1}(1)[/itex]. The range of [itex]sin^{-1}(x)[/itex] is [itex][-\frac{π}{2},\frac{π}{2}][/itex].

EDIT: While trigonometric substitution is a valid way to solve this definite integral, you can also solve the problem by thinking about what the graph of [itex]y=\sqrt{4-(x-3)^{2}}[/itex] is, and remembering that a definite integral is the area between the curve and the x-axis.

I know. I can solve this integral by modelling it as a semi circle and find the area.

But here, I am curious about the trigonometric substitution.
 
  • #6
94
1
I'm not sure "why" exactly. But if I had to guess I'd say just because. That's the way the inverse sine function was defined. It was given a restricted range. I'm not sure you need a reason why since it's in the definition of the inverse sine function.

EDIT: You can use [itex]\frac{3π}{2}[/itex] as your lower limit of integration and [itex]\frac{5π}{2}[/itex] as your upper limit if you want. You'll get the same answer since [itex]4cos^{2}u[/itex], the integrand you end up with after the substitution, has a period of [itex]π[/itex].
 
Last edited:
  • #7
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722

Homework Statement



[itex]\int_{1}^{5}\sqrt{4-(x-3)^2}dx[/itex]

Homework Equations





The Attempt at a Solution


Not really a homework. Just curious.

Let [itex]x-3=2sin u[/itex]
[itex]x=2sin u+3[/itex]

Now limit change,
when x=1, u = arcsin (-1)
when x=5, u = arcsin (1)

Since there are infinitely many solution of arcsin (-1) and arcsin (1), How do we know which ones to choose?

Thank you very much

When you change variables to [itex] x = 3 + 2 \sin(t)[/itex] you can see directly that [itex]x[/itex] ranges from 1 to 5 when [itex]t[/itex] ranges from [itex] -\pi/2 \text{ to } \pi/2. [/itex] We don't need to bother with different branches of the arcsin function; in fact, we can forget about arcsin completely----just look at the original requirements.

BTW: it is perfectly OK to use "[tex ]" and "[/tex ]" (no spaces); that will give you a displayed result, like [tex] \int_1^5 \sqrt{4 - (x-3)^2} \, dx, [/tex] instead of the in-line result you get when using "[itex ]" and "[/itex ]", which would write the same expression as [itex] \int_1^5 \sqrt{4 - (x-3)^2} \, dx.[/itex]

RGV
 
  • #8
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,365
1,033
...

Now limit change,
when x=1, u = arcsin (-1)
when x=5, u = arcsin (1)

Since there are infinitely many solution of arcsin (-1) and arcsin (1), How do we know which ones to choose?

Thank you very much
BTW, there are not infinitely many solutions for arcsin(-1) and arcsin(1).

The arcsine function is a perfectly well-defined function for which arcsin(-1) = -π/2 and arcsin(1) = π/2 .

However, each of the following two equations do have infinitely many solutions.
[itex]1=2\sin(u)+3[/itex]

[itex]5=2\sin(u)+3[/itex]​
As Ray V. mentioned, you don't need to consider all of those solutions for this problem.
 
  • #9
Alright. Thank you so much, you guys.
 

Related Threads on Substituting Integration with trigonometric function

Replies
5
Views
3K
Replies
2
Views
2K
Replies
3
Views
1K
Replies
4
Views
1K
Replies
2
Views
2K
Replies
3
Views
3K
Replies
9
Views
2K
Replies
4
Views
2K
Top