# Thread: Integration using... Inverse trig sub? or Solving the square?

1. ## Integration using... Inverse trig sub? or Solving the square?

$\displaystyle \int \sqrt{4 x-x^2} dx$
Okay so I've ran into another bit of a snag...
I know I can solve by the square and ill get
sqrt(4-(x-2)^(2))

but then again the question looks alot like the derivative of arcsin, which can be changed into arcsin...

What would be best fastest most efficient way of getting there?

2. Originally Posted by Lolcats
$\displaystyle \int \sqrt{4 x-x^2} dx$
Okay so I've ran into another bit of a snag...
I know I can solve by the square and ill get
sqrt(4-(x-2)^(2))

but then again the question looks alot like the derivative of arcsin, which can be changed into arcsin...

What would be best fastest most efficient way of getting there?
For:
$\displaystyle \int \sqrt{4-(x-2)^2} \,\ dx$.
Let $\displaystyle x-2=2sin(\theta)d\theta$.
then $\displaystyle dx = 2cos(\theta)d\theta$.
$\displaystyle \int .... = \int \sqrt{4-4sin^2(\theta)} \,\ 2cos(\theta)d\theta$
$\displaystyle =4\int cos^2(\theta)d\theta$.
Can you take it from here ?

3. Yes. Thank you very much.
However, I'm wondering when your doing that kind of substitution, how do you know that x-2=2sin(u) is equivalent to one another?

I understand that sin(u) would equal x-2 / 2, but where would you get the 2 from to begin with?

Do you assign 2 from x-2 as "a"?

4. Originally Posted by Lolcats
Yes. Thank you very much.
However, I'm wondering when your doing that kind of substitution, how do you know that x-2=2sin(u) is equivalent to one another?

I understand that sin(u) would equal x-2 / 2, but where would you get the 2 from to begin with?

Do you assign 2 from x-2 as "a"?
Do you the trigonometric substitution?

5. Shouldn't the answer also be 2x+2sin2x?

6. Somewhat, not really, the way my prof went over it just confused me, do you know of a link to a website thats really good at explaining it?

7. Originally Posted by Lolcats
Somewhat, not really, the way my prof went over it just confused me, do you know of a link to a website thats really good at explaining it?
8. Integration by Trigonometric Substitution

8. Wow, okay that makes a ton of sense. Thank you SO SO much.

9. You say that it looked to you like "arcsine". Notice that you could have used either x-1= 2cos(u) or x-1= 2sin(u)- they both give the same result.

You could also argue this way: Factor the "4" out: $\displaystyle \sqrt{4- (x-1)^2}= 2\sqrt{1- \frac{(x-1)^2}{4}}$ which should remind you of $\displaystyle \sqrt{1- cos^2(\theta)}= \sqrt{sin^2(\theta)}= sin(\theta)$ or $\displaystyle \sqrt{1- sin^2(\theta)}= \sqrt{cos^2(\theta)}= cos(\theta)$ so that you can use either $\displaystyle \frac{x-1}{2}= sin(\theta)$ or $\displaystyle \frac{x-1}{2}= cos(\theta)$ to clear the square root.