Question

Integral sqrt ( 1 - x^2) dx, from -1 to 1.

Answer 1

try trig substitution:
x=sin(u)
dx=cos (u) du

let me know if there is a problem

Answer 2

I sqrt ( 1 - sin^2 u) cos u du =
I cos u cos u du =
I cos^2 u du

And then?

Answer 3

cos^2x = 1/2 (1+cos2x)
and take integral of each term

Answer 4

I got:
= 1/2 u +1/4 sin2u

Answer 5

Ok.

I cos^2 u du =
1/2 I 1 + cos 2 u du
1/2 * u + 1/2 sin 2 u / 2 =
1/2 u + 1/4 sin 2u

Answer 6

now, what it the u=...?
x=sin u
then u=arcsin x
right?

Answer 7

sin2u=2sin u * cos u = \[=2x*\sqrt{1-x^{2}}\]

Answer 8

apply your (-1,1) to the final answer

Answer 9

The, so I can work straight with u, I'll change - 1 to 1 to sin u = - 1 to sin u = 1
u = 270° and 90°.

1/2 * 3 pi /2 + 1/4 * 0 - ( 1/2 * pi/2 + 1/4 * 0 ) =
3 pi / 4 - 1/4 * pi =
2/4 pi = 1/2 pi

Answer 10

should work!
is your answer matched?

Answer 11

No wait a minute, sin pi = 1. So:
3 pi /4 - (1/4 pi + 1/4) =
2/4 pi + 1/4....

Answer 12

2/4 pi - 1/4.

Answer 13

It's wrong though, the answer should be pi.

Answer 14

I didn't calculate... should I?

Answer 15

Yes.

Answer 16

give me a min

Answer 17

Ok.

Answer 18

i got -pi/2....?

Answer 19

Ah, you know what? Let's just leave it aside for a moment. Thanks for the help, sometime I'll still get round to do it.

Answer 20

OK. what it the answer should be - I'll play with it :)

Answer 21

The answer should be pi, see, this integral gives the area of a r=1 circle.

Answer 22

oooo....
because it's sq rt - graph is symmetrical over y-axis!

Answer 23

it should be :
\[=2*\int\limits_{0}^{\pi/2}\cos ^{2}u du=....\]
=2(1/2u+1/4 sin2u)=u + 1/2 sin2u=
=pi +0=pi
Yes!