Question

Solve this integral...

Answer 1

Use a substitution \(u=1-x^2\).

Answer 2

its not working here.. i've already tried

Answer 3

substitute 1-u^2=a and multiply 2 on top so u get the direct form

Answer 4

We have the integral

\(\large I=-\frac{1}{2} \int -2x(1-x^2)^{-\frac{1}{2}}dx \). Substitute \(u=1-x^2 \implies du=-2xdx\), the integral becomes:
\[-\frac{1}{2}\int u^{-\frac{1}{2}}du=\dots\]

Answer 5

ya thats the way ---the easiest one

Answer 6

let me know the final answer?

Answer 7

No, seriously, you should be able to calculate it from here. What do you get?

Answer 8

sorry its (-1)*((1-x^2)^(1/2))

Answer 9

yep

Answer 10

Thanks James for correcting me

Answer 11

explain meeeeee... i am not getting answerrr...:(

Answer 12

Can't you integrate \(u^{-\frac{1}{2}}\)?

Answer 13

=−1/2∫−2x/(1−x^2)^(−1/2)dx. Substitute u=1−x^2du=−2xdx, the integral becomes:
−1/2∫u^(−1/2)du=… this is =( u^(1/2))/((1/2)) *(-1/2)=-1*u^(1/2)....substitute u and get the answer

Answer 14

from where -1/2 comes in the beginning?

Answer 15

i multiplied the numerator with -2 and divided it with -1/2 to make it the same in order to get -2X in the numerator

Answer 16

Jetly also tell me we have this integral −1/2∫−2x/(1−x^2)^(−1/2)dx but after substitution it becomes −1/2∫u^(−1/2)du, where does -2x gone?

Answer 17

Oh i got it.. Thanks 2 awl...:)