What is the indefinite integral of y=(1-(x^2))^(1/2)?

Question

Please give reasoning for your answer.

Answer 1

y=(1-x^2)^(1/2)

here ,we use the substitution
x=sinp,then dx=cospdp
1-x^2=1-sin^2(p)=cos^2(p)
cosp=(1-x^2)^(1/2)

int(1-x^2)^(1/2)dp
=int(cosp*cosp)^(1/2)dp
{plugging in substitution}
=1/2int(1+cos2p)dp
{trig identity}
=1/2(p+1/2sin2p)
=1/2(p+sinp*cosp)
{trig identity}
but,cosp=(1-x^2)^(1/2),
p=sin^-1(x)
{above}
=1/2sin^-1(x)
+1/2x*(1-x^2)^(1/2)
=1/2{arcsinx+x*(1-x^2)^(1/2)}
+C

where C is an arbitrary constant

if there is anything you don't
follow,give me a call-this is
an important substitution

i hope that this helps

Answer 2

y' = ∫ (1 - x^2)^(1/2) dx

Let x = sin(u)
Thus, (1 - x^2)^(1/2) = [1 - sin^2(u)]^(1/2)

But, 1 - sin^2(u) = cos^2(u)
from the standard formula : sin^2(u) + cos^2(u) = 1

Therefore, (1 - x^2)^(1/2) = [cos^2(u)]^(1/2) = cos(u)
And from x = sin(u), we get dx = cos(u) du.

Substituting then gives :
y' = ∫ cos(u) * cos(u) du = ∫ cos^2(u) du

From another standard formula :
cos(2u) = 2cos^2(u) - 1 ,
we get :
cos^2(u) = (1/2) [cos(2u) + 1]

Now we have :
y' = ∫ (1/2)[cos(2u) + 1] du

= (1/2) ∫ [cos(2u) + 1] du

= (1/2) ∫ cos(2u) du + (1/2) ∫ du

y = (1/2)[sin(2u) / 2] + u / 2 , [because ∫ cos(ax) dx = sin(ax) / a]
= sin(2u) / 4 + u / 2
= sin(u)*cos(u) / 2 + u / 2 [because sin(2u) = 2sin(u)cos(u)]

Now we need to substitute values for u, sin(u) and cos(u).
From the initial x = sin(u), we get : u = sin^-1(x)
and we also have (1 - x^2)^(1/2) = cos(u).

Therefore, y = x(1 - x^2)^(1/2) / 2 + sin^-1(x) / 2

Answer 3

(1-x^2)^1/2
tis is a standard integral
=x(1-x^2)^1/2+(1/2)sin^-1x+C

Answer 4

Let x=sint => dx/dt=cost

y=(1-(sint)^2)^(1/2)
y=cost

dy/dt = -sint

dy/dx
= dy/dt x dt/dx
= -sint/cost
= -sint/(1-(sint)^2)^(1/2)
= -x/(1-x^2)^(1/2)