integration of the form (x^2)/(1-(x^2))^0.5?

Question

could anyone please help with the general integration of a function in the form of: (x^2)/(1-(x^2))^0.5

ie (x^2)/sqrt(1-(x^2))

Answer 1

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

To solve this, you use trigonometric substitution.
Let x = sin(t). Then
dx = cos(t) dt, so we have

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

Now, use the half angle identity, which goes

sin^2(z) = (1/2) (1 - cos2z)

Integral ( (1/2) (1 - cos2t) dt )

Pull out the constant (1/2),

(1/2) * Integral [(1 - cos2t) dt]

Now, this is more easily integrable (since 2t inside the cos is linear, and linear functions need only be offset by a constant when integrated).

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

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

Now, we use the double angle identity sin(2t) = 2sin(t)cos(t)

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

To express our answer in terms of x, note that since we let
x = sin(t), it follows that
sin(t) = x/1.

Note that by the trig rules with right angle triangles,
sin(t) = opp/hyp. Therefore
opp = x
hyp = 1, and by Pythagoras,
adj = sqrt(hyp^2 - opp^2) = sqrt(1^2 - x^1) = sqrt(1 - x^2).

For (1/2)t - (1/2)sin(t)cos(t) + C,

sin(t) = opp/hyp = x/1 = x
cos(t) = adj/hyp = sqrt(1 - x^2) / 1 = sqrt(1 - x^2).
Also, since x = sin(t), then t = arcsin(x), so our integral is

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

Answer 2

let x=sinQ
dx=cosQdQ
change x=sinQ when you see x

int(sin^2Q)*cosQdQ/sqrt(1-sin^2Q)
but 1-sin^2Q=cos^2Q
int(sin^2Q)*cosQdQ/sqrt(cos^2Q)
but sqrt(cos^2Q)=cosQ
so int (sin^2Q)*cosQdQ/cosQ
int(sin^2Q)dQ

but sin^2Q=(1-cos2Q)/2
int((1-cos2Q)/2 dQ
int 1/2 dQ -intcos2Q/2dQ
=1/2 Q -1/4 sin2Q +C
but x=sinQ=>Q=sin^-1(x)
but it back into it
(1/2)*sin^-1(x)-1/4*sin(2*sin^-1(x))+C
int=integral, ^ square sin^-1x is inverse sinx * time
answer is
1/2)*sin^-1(x)-1/4*sin(2*sin^-1(x))+C
sorry for spacing after all the dot will follow this one at the first we have 1-sin^2(x)..)
second cos^2Q)
third 2*sin^-1(x))+C
fourth 2*sin^-1(x))+C
good luck

Answer 3

u = x^2 dv/dx = (1-x^2)^-0.5

du/dx = 2x , v = 1/x * (1-x^2)^0.5

I = x(1-x^2)^0.5 - (I)2(1-x^2)^0.5

= x(1-x^2)^0.5 - 2(1-x^2)^1.5/(3x) +C

Answer 4

put x=sina
therefore dx=cosa.da
integral becomes
sin^2a.cosa.da/cosa
=sin^2a.da
=(1-cos2a)/2.da
=a/2+(sin2a)/4
=(sin^-1)/2+{sin(2sin^-1a)}/4