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thanks for any help.

btw, i'm supposed to turn it in by midnight. any help before then is appreciated

might be easier to write it out then scan it and send it.

2007-11-21 14:15:02 · 5 answers · asked by bob oso 2 in Science & Mathematics Mathematics

5 answers

∫4(9-x²)^(1/2) dx

Let x=3sin(u)
dx=3cos(u)du

sin(u) = x/3
u = arcsin(x/3)

∫4(9-x²)^(1/2) dx
= ∫4(9-9sin²u)^(1/2)*3cos(u)du
= ∫12(9cos²u)^(1/2).cos(u)du
= ∫12(3cos(u))cos(u)du
= ∫36cos²udu

cos(2u) = 2cos²u-1
cos²u = (1/2)(1+cos(2u))
36cos²u = 18(1+cos(2u))

∫36cos²udu
= ∫18(1+cos(2u))du
= 18u + 9sin(2u) + c
= 18u + 18sin(u)cos(u) + c
= 18arcsin(x/3) + 18/3(x/3)(9-x²)^(1/2) + c
= 18arcsin(x/3) + 2x(9-x²)^(1/2) + c

You can check the answer on the attached link.

2007-11-21 14:51:44 · answer #1 · answered by gudspeling 7 · 0 0

You have to integrate the square root of a quadratic
function. You want to make a trig substitution which
will eliminate the radical.
So look at √(9-x²). If we let x = 3 sin t
then √(9-x²) = √(9-9 sin² t) = 3√(1- sin² t) = 3 cos t.
Also dx = 3 cos t dt.
So we have to integrate
36∫ cos² t dt
Now recall that cos² t - sin² t = cos 2t
and cos² t + sin² t = 1.
If we add these 2 equations, our integral becomes
18∫ (1 + cos 2t dt) = 18(t + ½ sin 2t ) + C =
18(t + sin t cos t) + C.
Now back substitute to get the final answer:
18(arcsin x/3 + x/3*√(1 - x²/9) ) + C
= 18 arcsin x/3 + 2 x*√(9-x²) + C.

2007-11-21 23:11:46 · answer #2 · answered by steiner1745 7 · 0 0

What trig functions do you know that satisfy ( )^2 + ( )^2 = constant...only two. So try 3 times one of those functions. And remember that cos(2x) = 1 - 2(sin x)^2=2 (cos x)^2 -1 You will need this identity.

2007-11-21 22:50:32 · answer #3 · answered by ted s 7 · 0 0

4(9 - x^2)^1/2 = 4 square root [ (3^2 - x^2 )]
we have
integrate 4(square root[(3^2 - x^2)]dx
, use x = 3sin(t) , INTERCHANGE
we have
intergrate 4 (square root (3^2 - (3sin(t)^2)]dt
INTEGRATE 4 (SQUARE root [(3^2) [ 1 - sin(t)^2] dt
= 12 (INTEGRATE square root ( 1 - sin(t) ^2 ) dt
BUT COS(t)^2 = 1 - SIN(t)^2 , SO
=12 (INtegrate square root ( cos(t)^2 )dt
= 12 ( INtegrate (cos(t) dt
= 12 SIN(t) + C

2007-11-22 00:48:32 · answer #4 · answered by LE THANH TAM 5 · 0 1

4x9 is 36
4x^2 is 16x^1/2 is 8x then divide 36 by 8 on both sides 8 cancels out ur left with x so x equals 4.5

good luck hope i am right

2007-11-21 22:26:25 · answer #5 · answered by spicy delicious 2 · 0 3

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