# Substituting Integration with trigonometric function

1. May 20, 2012

### inter060708

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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

Last edited: May 20, 2012
2. May 20, 2012

### 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.

3. May 20, 2012

### 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.

Last edited: May 20, 2012
4. May 20, 2012

### inter060708

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

5. May 20, 2012

### inter060708

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. May 20, 2012

### 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 $π$.

Last edited: May 20, 2012
7. May 20, 2012

### Ray Vickson

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

8. May 20, 2012

### SammyS

Staff Emeritus
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.

9. May 21, 2012

### inter060708

Alright. Thank you so much, you guys.