Question

integrate 2x/sqrt(1-x^2) dx from 0 to 1/2

Answer 1

try a substitution
\[\text{let } u=1-x^2\]

Answer 2

alrighty

Answer 3

kids ok? :)

Answer 4

haha yes, they are

Answer 5

im stuck.

Answer 6

i have \[-1/2\int\limits_{1}^{3/4}du/\sqrt{u} du\]

Answer 7

i dont think the top of the function is right, it was the value i had for du but idk.....

Answer 8

I would do trig substitution here.

Answer 9

what do you mean?

Answer 10

\[
\int\frac{2x}{\sqrt{1-x^2}}dx\\
x=\sin u\\
dx=\cos u\ du\\
\int\frac{2\sin u\cos u}{\sqrt{1-\sin^2u}}du=\int\frac{2\sin u\cos u}{\sqrt{\cos^2u}}du=\int\frac{2\sin u\cos u}{\cos u}du=\int2\sin u\ du
\]

Answer 11

how would you know to use that, since there is no trig functions in the equation.

Answer 12

Any equation with a denominator like that is often a good candidate for a trig substitution. You'll be able to recognize them after enough practice.

Answer 13

it looks like inverse sin

Answer 14

@nbouscal it's very simple integral
She just forgot to take root of the upper limit, it should be sqrt (3) /2

Answer 15

much easier to do u=1-x^2

Answer 16

Eh, there are plenty of methods to do any given integral. I find trig substitution easiest because I can usually do it in my head.

Answer 17

the derivaitve if available so i second that motion^

Answer 18

\[u=1-x^2 => du=-2x dx => -du =2 x dx\]
\[\int\limits_{1-0^2}^{1-(\frac{1}{2})^2} \frac{-du}{\sqrt{u}}\]
\[\int\limits_{1}^{\frac{4}{4}-\frac{1}{4}}-u^\frac{-1}{2} du\]
\[-\int\limits_{1}^{\frac{3}{4}}u^{-\frac{1}{2}} du\]

Answer 19

Yeah, I guess it's pretty easy that way too. Meh. Either way works fine.

Answer 20

I wasn't saying your way didn't work. I was just saying the way I chose is much easier

Answer 21

Matter of opinion.

Answer 22

would there be a -1/2 in front of the integral?

Answer 23

lol ok. You learn trig sub after algebraic sub
But okay
I guess it is sorta an opinion thing

Answer 24

@jackiefml what does du equal?
du=-2x dx
so
-du=2x dx
Isn't 2x dx on top
So isn't that -du?

Answer 25

yes, but normally i have to separate the coeffecient from the dx.

Answer 26

ok let me know if you run into anymore trouble

Answer 27

i integrate next to get rid of the du?

Answer 28

This symbol in the picture means you need to integrate with respect to u
yes

Answer 29

ok so id get.... \[2\sqrt{u}\]

Answer 30

don't forget the negative

Answer 31

ok

Answer 32

Then evaluate your antiderivative at the upper limit - evaluate it at the lower limit

Answer 33

and if you prefer @nbouscal 's way then look at the attachment
I did it both ways