how do i integrate x*sqrt(1-x^2)?

Answer 1

here's an easy way

int{x*sqrt(1-x^2)dx}

let u=1-x^2.....(1)
then,du/dx= -2x
giving dx= -du/2x

int{x*sqrt(1-x^2)dx}
=int{x*sqrt(u)*(-du/2x)}
= (-1/2)*int(sqrt(u))du
= -(1/2)*u^(3/2)/(3/2) +C
= -(1/2)*(2/3)*u^(3/2)+C
= -(1/3)*(u)^(3/2)+C
but,from (1), u=1-x^2

therefore,substituting for u,

int{x*sqrt(1-x^2)dx}
= -(1/3)*(1-x^2)^(3/2)+C

where C is an arbitrary constant

i hope that this helps

Answer 2

this can be done with the clever substitution u = x^2, then du = 2dx
with gives :

Int x*sqrt(1-x^2) dx = Int 1/2 sqrt(1-u) du

next set v = 1-u then dv = -du, so you get

Int -1/2 sqrt(v) dv and that is solvable..

Answer 3

suppose 1-x^2=t
differentiating it
that is 0-2x=dt
2x=-dt
x=-1/2 dt
substitute in x*sqrt of 1-x^2
you will get 1/2 *sqrt of t dt
then integrating it you will get
1/3 * t^3/2
then sub t
that is 1/3 * (sqrt of 1-x^2)^3/2

Answer 4

I wonder what you did to deserve those answers?

This is a u-substitution problem. Let u=1-x^2, then du=-2x dx

You don't have -2x dx but can rewrite to get there:
-1/2 ∫√(1-x^2) -2x dx

Now substitute:
-1/2 ∫ √u du

This you can integrate. Then, back-substitute to get a function of x

Answer 5

Let by y=sqrt(1-x^2) , then y^2=1-x^2 => 2ydy=-2xdx /:2 =>ydy=-xdx, then you has int=-int(y*y)dy=-int(y^2)dy=-1/3y^3=-1/3(1-x^2)sqrt(1-x^2)+C . END!!

Answer 6

set 1-x^2=t
-2xdx=dt
xdx=-(1/2)dt
now the integral=(-1/2)t^-1/2dt
=(-1/2)*t^1/2/1/2
=-(1-x^2)^1/2+C

Answer 7

easy multiply your date of birth divide by the size of your feet subtract your next door neighbours age, result

Answer 8

it is easy.use x=sint or x=cost.dx=costdt
calculate:

Answer 9

Forget what I said before, that was for derivating...

Answer 10

easy, use TI-89.