# Homework Help: Another tri integration

1. Apr 7, 2010

### jwxie

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

Integral of 6/(sqrt(4-x^2))

2. Relevant equations

arcsin(u) = u'/ sqrt(1-u^2)

3. The attempt at a solution

WA gave a different answer
http://www.wolframalpha.com/input/?i=Integral+of+6%2F%28sqrt%284-x^2%29%29

I got the following:

Integral of 6/(sqrt(4-x^2))
divide by 4 to reduce to 1-u^2, and pull out the constant 4: 1/4 int of 6/sqrt(1-(x/2)^2
i was going to pull out the six right away, but i could integrate that with du.
u = x/2, du = 1/2 dx, so 2 du = dx
i get 12 inside, and pull out the 12, i get 12/4 int u' / sqrt(1-u^2)
which gives me 3 arcsin(x/2)+c

but this is not right from WA.

Thank you.

2. Apr 7, 2010

### tt2348

U sub
x=2sin(u),
dx=2cos(u)du.
It'll become
6*2cos(u)/((4-4(sin(u))^2)^(1/2)

3. Apr 7, 2010

### Staff: Mentor

$$\sqrt{4 - x^2} = \sqrt{4(1 - x^2/4)} = 2 \sqrt{1 - (x/2)^2}$$

I got the same as WA. This is a very simple trig substitution. I always draw a right triangle, since I don't like to clutter up my head with a bunch of formulas if I don't have to. The hypotenuse is 2, the opposite side is x, and the adjacent side is sqrt(4 - x^2).

From this I get 2sinw = x, and 2cosw dw = dx, and 2 cos w = sqrt(4 - x^2)

$$\int \frac{6 dx}{\sqrt{4 - x^2}} = 6 \int \frac{2cos w dw}{2cos w} = 6\int dw = 6w + C$$

From 2sin w = x, I get sin w = x/2 ==> w = sin-1(x/2), so the antiderivative is 6sin-1(x/2) + C.

YOU KEEP FORGETTING YOUR DX! IT WILL COME AROUND TO BITE YOU ONE DAY!

4. Apr 7, 2010

### jwxie

Hi. Thank you for doing these for me.
I think I have learned the mistake.
All the sub are corected except when I pulled out the 1/4 , i didn't square root it. It should be 1/2 outside.

LOL Thanks. I just left dx out since there was only one variable, namely, x.

But I WILL REMEMBER THAT...

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook