Question

one simple integral!
x/sq.rt(1-x^4)??

Answer 1

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

Answer 2

let x^2 = y

Answer 3

then use trig subs.

Answer 4

i did. u=x^2 du=2xdx
\[\frac{ 1 }{ 2 }\int\limits_{-1/2}^{1/2}\frac{ du }{ \sqrt{1-u^2} }=\sin^{-1}u \]

Answer 5

is this correct?

Answer 6

Method 1: Note that \(\large\displaystyle \int_{-1/\sqrt{2}}^{1/\sqrt{2}} \frac{x}{\sqrt{1-x^4}}\,dx = \int_{-1/\sqrt{2}}^{1/\sqrt{2}} \frac{x}{\sqrt{1-(x^2)^2}}\,dx\). Let \(\large u=x^2\) and recall that \(\large\displaystyle\int \frac{1}{\sqrt{1-y^2}}\,dy = \arcsin y + C\)

Method 2: Note that \(\large \dfrac{x}{\sqrt{1-x^4}}\) is an odd function. Do you recall what the integral of an odd function over a symmetric interval \(\large [-a,a]\) is?

Answer 7

@emilyoo the lower limit of your new integral isn't correct. Note that \(\large \left(-\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{1}{2}\).

Answer 8

oh i see i made a mistake

Answer 9

thank you!

Answer 10

No problem. So what did you get in the end? :-)

Answer 11

0

Answer 12

Good. :-)