use trigonometric substition to integrate x^(2)(1-x^(2))^(1/2)dx?

Answer 1

Use the u = sinx or cosx substitution.
For u = sinx:
INT(sin^2(x)*(1-sin^2(x))^(1/2))dx
using substitution of 1 = sin^2(x) + cos^2(x)
INT(sin^2(x)*(cosx))dx
now use another substitution:
v = sinx and dv = cosxdx
INT(v^2 * dv)
integrate with respect to v now:
(1/3)*v^3
and substitute for v
(1/3)*sin^3x
If this is a definite integral you dont have to change the limits a and b because we substituted back into the x-y plane.

Answer 2

You can consider x=siny then you shoul substitude the integral by new parameter,
x^2(1-x^2)dx=sin^2 (y) [1-cos^2 (y)]^1/2dx= sin^2 (y) [1-cos^2 (y)]^1/2(cosy dy)= sin^2 (y) cos^2 y dy,
so you can integrate the final states:
sin^2 (y) cos^2 y dy
while y=Arcsinx

Answer 3

You will find that either of the substiutions x = sinA or x = cosA will reduce the integral to one that you should be able to do without too much difficulty. The first probably makes it a little easier. You don't really want me to give you every line leading to the answer, do you?

Answer 4

Integral ( x^2 (1 - x^2)^(1/2) dx )

Let x = sin(t).
dx = cos(t) dt.

Integral ( sin^2(t) (1 - sin^2(t))^(1/2) cos(t) dt )

Integral ( sin^2(t) (cos^2(t))^(1/2) cos(t) dt)

Integral ( sin^2(t) cos(t) cos(t) dt )

Integral ( sin^2(t) cos^2(t) dt)


Now we have to use the half angle identity to solve this.
sin^2(t) = (1/2)(1 - cos(2t))
cos^2(t) = (1/2)(1 + cos(2t))

Integral ( (1/2)(1 - cos(2t)) (1/2)(1 + cos(2t)) dt )

(1/4) Integral ( (1 - cos(2t))(1 + cos(2t)) dt )

(1/4) Integral ( ( 1 - cos^2(2t) ) dt )

Use the half angle identity again, on cos^2(2t).

(1/4) Integral ( (1 - (1/2)(1 + cos(4t)) ) dt )

(1/4) (t - (1/2)(t + (1/4)sin(4t) ) ) + C

(1/4) (t - (1/2)t - (1/8)sin(4t) ) + C

(1/4)t - (1/8)t - (1/32)sin(4t) + C

(1/8)t - (1/32)sin(4t) + C

Use the double angle identity;
sin(4t) = 2sin(2t)cos(2t).

(1/8)t - (1/32)[2sin(2t)cos(2t)] + C

(1/8)t - (1/16)sin(2t)cos(2t) + C

(1/8)t - (1/16) 2sin(t)cos(t)[cos^2(t) - sin^2(t)] + C

To solve back in terms of x, remember that
x = sin(t), so
sin(t) = x/1 = opp/hyp
opp = x
hyp = 1, so by Pythagoras,
adj = sqrt(1 - x^2)

sin(t) = x/1 = x
cos(t) = adj/hyp = sqrt(1 - x^2)/1 = sqrt(1 - x^2)

(1/8)t - (1/8) sin(t)cos(t)[cos^2(t) - sin^2(t)] + C becomes

(1/8) arcsin(x) - x sqrt(1 - x^2) [1 - x^2 - x^2] + C

(1/8) arcsin(x) - x sqrt(1 - x^2) [1 - 2x^2] + C