Question

Integral HELP!

Answer 1

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

Answer 2

Does anyone have any idea how to solve it?

Answer 3

@SolomonZelman ?

Answer 4

\(\large\color{slate}{\displaystyle\int\limits_{0}^{1/ \sqrt{~2}}\frac{x}{\sqrt{1-x^4}}~dx}\)

\(\large\color{slate}{\displaystyle\int\limits_{0}^{1/ \sqrt{~2}}\frac{x}{\sqrt{1^2-(x^2)^2}}~dx}\)

\(\large\color{slate}{\displaystyle\int\limits_{0}^{1/ \sqrt{~2}}\frac{x}{\sqrt{(1^2-x^2)(1^2+x^2)}}~dx}\)

\(\large\color{slate}{\displaystyle\int\limits_{0}^{1/ \sqrt{~2}}\frac{x}{\sqrt{(1-x^2)(1+x^2)}}~dx}\)

Answer 5

does this make sense so far?

Answer 6

I used the difference of squares.
(if you don't get it, ask)

Answer 7

yes I tried that

Answer 8

(the next step would be u=x^2)

Answer 9

I got there so far

Answer 10

\(\large\color{slate}{\displaystyle\int\limits_{0}^{1/ \sqrt{~2}}\frac{x}{\sqrt{(1-x^2)(1+x^2)}}~dx}\)

\(\large\color{slate}{\displaystyle u=x^2}\)
\(\large\color{slate}{\displaystyle du=2x~dx}\) --> \(\large\color{slate}{\displaystyle (1/2)~du=x~dx}\)

\(\large\color{slate}{\displaystyle x=1/\sqrt{2}}\) --> \(\large\color{slate}{\displaystyle u=1/2}\)
\(\large\color{slate}{\displaystyle x=0}\) --> \(\large\color{slate}{\displaystyle u=0}\)

\(\large\color{slate}{\displaystyle\int\limits_{0}^{1/ 2}\frac{1}{2}\frac{1}{\sqrt{(1-u)(1+u)}}~du}\)

Answer 11

arctg!

Answer 12

what , sorry ?

Answer 13

\(\large\color{slate}{\displaystyle\frac{1}{2}\int\limits_{0}^{1/ 2}\frac{1}{\sqrt{1-u^2}}~du}\)

Answer 14

sorry my bad, I thought the solution for this integral is arctang

Answer 15

arcsinu right?

Answer 16

yup

Answer 17

\(\large\color{slate}{\displaystyle\frac{1}{2}\int\limits_{0}^{1/ 2}\frac{1}{\sqrt{1-u^2}}~du~~=~~\frac{1}{2}\sin^{-1}(u)~{\Huge|}^{1/2}_{0}}\)

Answer 18

then plug in the limits of integration

Answer 19

thanks again!

Answer 20

I got the right answer thanks to you!

Answer 21

you can use http://www.wolframalpha.com/
for your calculations

Answer 22

yw

Answer 23

I shall

Answer 24

I will depart for 1 hour or so..... bye