Question

I need help with some Integration.

Answer 1

\[\int\limits_{}^{} xdx/(\sqrt(1-x^2))\]

Answer 2

I know that 1/sqrt(1-x^2) is sin^-1(x) but im not sure how to get it in that form.

Answer 3

Looks like something where substitution would help.

Answer 4

Take 1-x^2 = t^2 and try

Answer 5

so i shouldnt use the trig identitiy at all here

Answer 6

Nope . It is too easy with the substitution I gave above

Answer 7

Take 1-x^2 = t^2
So
-xdx = tdt.

Substitute both in your integration problem.You should get the answer

Answer 8

I couldn't make your substitution work for some reason, you could alternatively try x=sinu and I assure you that can be just as easy.

Answer 9

ya im having trouble with shivam_bhalla's substitution.

Answer 10

is there no way that i can do (x)(1/sqrt(1-x^2))(dx)

and take the integral using the trig identity?

Answer 11

\[\int\limits\frac{x}{\sqrt{1-x^{2}}}dx \]
\[u= 1-x^2 => \frac{du}{-2x} \]
\[\int\limits\frac{x}{\sqrt{u}} *\frac{du}{-2x} =>-\frac{1}{2} \int\limits\frac{1}{\sqrt{u}} \]
Then integrate it..
or you can alternatively use trig-substitution; \(x=sin\theta\)

Answer 12

im just a little confused on how you got du/-2x?

Answer 13

i realize you took the derivative of u. but why move it to the same side? and wouldn't it be positive 2x in the denominator

Answer 14

\(u= 1-x^2 => dx=\frac{du}{-2x}\)

Answer 15

ok i see

Answer 16

or you can use trig-sub \(x=cos\theta\)

Answer 17

so how did you pull out the -1/2? and no i dont think i wanna mess with trip substitution unless its and identity i can simply integrate haha

Answer 18

hm..the \(x\) cancels out so its \(-\frac{1}{2}\)