Question

\[Calculate: \int\limits_{0}^{1/2}4x/\sqrt{1-x^2}dx\]

Answer 1

I'm aware that I should substitute. The answer I come up with is - according to the book - not correct.

Answer 2

\[-4+4\sqrt{3/4}\] Is what I came up with.

Answer 3

try\[x ^{2} = t\]
and dt =2xdx

Answer 4

Trying!

Answer 5

\[2\int\limits_{0}^{1/4}dt/\sqrt{1-t}\]

Answer 6

I got -4+2*sqrt(3/4)
Which according to the book is wrong, did I go wrong somewhere?

Answer 7

what does the book say is correct?

Answer 8

4-2sqrt(3)

Answer 9

if anyone has doubt about answer you can always check at wolframalpha

Answer 10

I can't, don't know how to write the integrals! :c

Answer 11

i think my answer should be correct

Answer 12

let's ask wolf abt that .. hehe

Answer 13

"integral from 0 to 1/2 of 4x/sqrt(1-x^2) dx"
wolfram agrees with book

Answer 14

Blargh! :x

Answer 15

accessdenied should have post the link instead

Answer 16

i would try substituting u = 1 - x^2

Answer 17

That's what I originally did eigenschmeigen, did not work out. I think I got the same result I do now, not sure.

Answer 18

true
http://www.wolframalpha.com/input/?i=integral+from+0+to+1%2F2+of+4x%2Fsqrt%281-x%5E2%29+dx
i dont really use this much, forgot about that. :P

Answer 19

@eigenschmeigen you always point out that ... now i will never forget that.

Answer 20

integral that i wrote befor value is:
\[2[-2\sqrt{(1-t)}]_{0}^{1/4}=2(2-2\sqrt{3/4}) =4 -2\sqrt{3}\]

Answer 21

How did the -2 get before the sqrt?

Answer 22

That's where I'm failing..

Answer 23

\[\int\limits (1-t)^{-1/2}\]

Answer 24

and -t ofcource

Answer 25

Ah, brainfart, thanks alot!

Answer 26

ya, it's ugly

Answer 27

Thanks, myko! :D

Answer 28

:D

Answer 29

\[\int\limits^{1/2}_0{\frac{4x}{\sqrt{1-x^2}}dx}\]

\[u= 1-x^2\]
\[\frac{du}{dx}= -2x\]
\[\frac{-du}{2x}= dx\]

\[\int\limits^{1/2}_0{\frac{4x}{\sqrt{1-x^2}}dx}=\int\limits^{3/4}_1{\frac{-2}{\sqrt{u}}du}= -4[\sqrt{u}]^{3/4}_1 = 4 - 2\sqrt{3}\]

sorry if the question is already resolved - once i started i didnt want to stop

Answer 30

Don't be sorry - thanks for helping! I know what you mean by not wanting to stop, absolutely loving this chapter regarding substitution. Mathematically intoxicating. :3

Answer 31

absolutely agree, gotta love substitution