Substituting Integration with trigonometric function

[inter060708]

Homework Statement

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

Homework Equations

The Attempt at a Solution

Not really a homework. Just curious.

Let x-3=2sin u
x=2sin u+3

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

 

[cjc0117]

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

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.

 

[cjc0117]

There are not infinitely many solutions to sin^{-1}(-1) and sin^{-1}(1). The range of sin^{-1}(x) is [-\frac{π}{2},\frac{π}{2}].

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 y=\sqrt{4-(x-3)^{2}} is, and remembering that a definite integral is the area between the curve and the x-axis.

 

[inter060708]

There are not infinitely many solutions to sin^{-1}(-1) and sin^{-1}(1). The range of sin^{-1}(x) is [-\frac{π}{2},\frac{π}{2}].

Why is that? can't it be [3∏/2, 5∏/2]? or let's say [-∏/2 , 5∏/2] ?

 

[inter060708]

There are not infinitely many solutions to sin^{-1}(-1) and sin^{-1}(1). The range of sin^{-1}(x) is [-\frac{π}{2},\frac{π}{2}].

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 y=\sqrt{4-(x-3)^{2}} 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.

 

[cjc0117]

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 \frac{3π}{2} as your lower limit of integration and \frac{5π}{2} as your upper limit if you want. You'll get the same answer since 4cos^{2}u, the integrand you end up with after the substitution, has a period of π.

 

[Ray Vickson]

Homework Statement

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

Homework Equations

The Attempt at a Solution

Not really a homework. Just curious.

Let x-3=2sin u
x=2sin u+3

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 x = 3 + 2 \sin(t) you can see directly that x ranges from 1 to 5 when t ranges from -\pi/2 \text{ to } \pi/2. 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 \int_1^5 \sqrt{4 - (x-3)^2} \, dx, instead of the in-line result you get when using "[itex ]" and "[/itex ]", which would write the same expression as \int_1^5 \sqrt{4 - (x-3)^2} \, dx.

RGV

 

[SammyS]

...

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.

1=2\sin(u)+3

5=2\sin(u)+3​

As Ray V. mentioned, you don't need to consider all of those solutions for this problem.

 

[inter060708]

Alright. Thank you so much, you guys.