Trigonometric substitution integration HELP!!?

Question

from 0 to 1

integrate [dx / ((4-x)^2)^(3/2)]


I just have no idea how to do this. I asked my professor, but he was unable to help me. Could anybody help me with this problem please?

Thank you

Answer 1

∫ 1/(4-x)³ dx

= 1/(4-x)² * (-1/2) * (-1)

= 1/18 - 1/8

= -5/72

Answer 2

I = ∫ 1 / (4 - x)³ dx between lims 0 to 1
I = ∫ (4 - x) ^ (- 3) dx (lims. 0 to 1)
Let u = 4 - x
du = - dx
when x = 0, u = 4
when x = 1,u = 3
I = - ∫ u^(-3) du between 4 and 3
I = ∫ u^(- 3) du between 3 and 4
I = u^(-2) / (- 2) between 3 and 4
I = (- 1/2).(1 / u²) between 3 and 4
I = (- 1/2)[ 1/16 - 1/9]
I = (- 1/2)[ 9 /144 - 16/144
I = (1 / 2).(7 / 144)
I = 7 / 288

Answer 3

if I read the problem correctly the answer I ended with was 1/(4sqrt3).
remember with trig sub to make the triangle to identify your substitutions. whenever there is subtraction the first item is the Hypotenuse value squared so label your hyp.=2. then make x the value of the side adjacent to the angle so that x/2=cos#
that leaves the opposite side to be the sqrt[4-x^2]. I can attach an email of my steps if you like?

Answer 4

∫[dx / ((4-x)^2)^(3/2)]
let u = 4 - x
du = - dx
- ∫ du / u^3 =
1/(2u^2) =
. . . . . . . . . . 1
1/(2(4 - x)^2)| =
. . . . . . . . . . 0
1/(2*3^2) - 1/(2*4) =
(4 - 9) / 72 =
- 5/72

Answer 5

I = x³ / ?(one hundred forty four - x²) enable x = 12 sin ? dx = 12 cos ? d? subsequently, ? x³ dx / ?(one hundred forty four - x²) = ? (12 sin ?)³ (12 cos ? d?) / [?{one hundred forty four - (12 sin ?)²}] = 1728 ? sin³ ? d? You have been perfect until this element. enable us to proceed. ? x³ dx / ?(one hundred forty four - x²) = ? (12 sin ?)³ (12 cos ? d?) / [?{one hundred forty four - (12 sin ?)²}] = 1728 ? sin³ ? d? = 1728 ? (sin² ? * sin ?) d? = 1728 ? (a million - cos² ?) * sin ? d? = 1728 ? (sin ? - sin ? cos² ?) d? = 1728 [ - cos ? + (a million/3)cos³ ? ] + c wish this facilitates! J

Answer 6

i think you should let
sqrt(x)=2sin(theta)
then square both side
x=(2sin(theta))^2
hope this help