If the substitution √x= sin(y) is made in the intergrand of ∫[1/2,0](√x/√(1-x))dx, the resulting intergral is?

Question

Please explain as best as possible!!

Answer 1

Don't understand why you would but int over [ 0, arcsin √0.5] of { 2 sin² y dy}....x = sin² y...x = 0 means y = 0, x = 1/2 means y = arcsin√0.5, 1 - sin² y = cos ²y, dx = 2 sin y cos y dy

Answer 2

∫ [√x /√(1 - x)] dx =

let √x = siny →

y = arcsin(√x)

x = sin²y →

dx = 2 siny cosy dy

then, substituting, you get:

∫ [√x /√(1 - x)] dx = ∫ [siny /√(1 - sin²y)] 2 siny cosy dy =

∫ [siny /√(cos²y)] 2 siny cosy dy =

2 ∫ (siny /cosy) siny cosy dy =

cosy canceling out,

2 ∫ sin²y dy =

recall the half-angle identity sin²y = (1/2)[1 - cos(2y)]:

2 ∫ (1/2) [1 - cos(2y)] dy =

2 (1/2) ∫ [1 - cos(2y)] dy =

breaking it up,

∫ dy - ∫ cos(2y) dy =

y - (1/2)sin(2y) + C =

according to double-angle identities:

y - (1/2)(2 siny cosy) + C =

y - siny cosy + C

summing up,

siny = √x →

y = arcsin(√x)

thus cosy = √(1 - sin²y) = √(1 - √x²) = √(1 - x)

therefore, substituting back, you get:

y - siny cosy + C = arcsin(√x) - √x √(1 - x) + C =

arcsin(√x) - √[x(1 - x)] + C

that is the antiderivative

finally, evaluating the definite integral from 0 to (1/2), you get:

{ arcsin[√(1/2)] - √{(1/2)[1 - (1/2)]} } - { arcsin[√(0)] - √{(0)[1 - (0)]} } =

[arcsin(1/√2) - √(1/2)(1/2)] - 0 =

[arcsin(1/√2) - √(1/2)(1/2)] - 0 =

π/4 - 1/2 = (π - 2)/4

I hope it helps...

Bye!

Answer 3

∫√x/√1-x dx ....from 0 to 1/2

Substituting

   √x = siny

    x = sin^2 y

    dx = 2 siny cosy dy

Also

y = sin^(-1)√x

So limit changes like this

for x = 0

y=sin^-1(0)=0

for x =1/2

√x=1/√2

y = sin^(-1)(1/√ 2)=pi/4

Now the integral becomes

∫[siny/√(1-sin^2y)](2siny cosy dy).............from 0 to pi/4

=∫[ siny/cosy][2sin y cos y dy]  .....cosy gets cancelled

= 2∫ sin^2y dy

= 2/2∫ 2sin^2y dy

We know 2sin^y = 1- cos2y

using this

=  ÃƒÂ¢Ã‚ˆÂ«(1-cos2y)dy

=  [∫dy- ∫cos2y dy ]  .........from 0 to pi/4

= [ y - 1/2 sin2y] ............from 0 to pi/4

= [ pi/4 -  1/2 sin(pi/2) - 0 +0]

= [pi/4 -1/2]

= pi/4 - 1/2