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Another tri integration

  • Thread starter jwxie
  • Start date
  • #1
280
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Homework Statement



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

Homework Equations



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

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.
 

Answers and Replies

  • #2
144
0
U sub
x=2sin(u),
dx=2cos(u)du.
It'll become
6*2cos(u)/((4-4(sin(u))^2)^(1/2)
 
  • #3
33,521
5,199

Homework Statement



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

Homework Equations



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

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.
[tex]\sqrt{4 - x^2} = \sqrt{4(1 - x^2/4)} = 2 \sqrt{1 - (x/2)^2}[/tex]

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)

[tex]\int \frac{6 dx}{\sqrt{4 - x^2}} = 6 \int \frac{2cos w dw}{2cos w} = 6\int dw = 6w + C[/tex]

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
280
0
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...
 

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