Question

Integrate

Answer 1

\[\Huge \int\limits_{}^{} xsin^{-1}x\]

Answer 2

arcinsx has priority over x so..
\[\Large \sin^{-1}x \int\limits_{}^{} xdx-\int\limits_{}^{}\frac{1}{\sqrt{1-x^2}}.\int\limits_{}^{}xdx\]

Answer 3

\[\Large \frac{1}{2}(x^2\sin^{-1}x-\int\limits_{}^{}\frac{x^2}{\sqrt{1-x^2}})\]

Answer 4

now to solve that thing i was thinking about substituting x=sin theta so whole of it becomes tan theta sec theta which integrates to sec theta..but that is just a thought..i want the easiest way to overcome it.

Answer 5

you are absolutely right! :')

Answer 6

o.O ?

Answer 7

x^2/sqrt(1-x^2) hmm
do it like this
write it as -(-x^2/sqrt(1-x^2)) . Keep the outside - sign out of the integral.
Now (-x^2/sqrt(1-x^2)) = (1-x^2 -1)/sqrt(1-x^2)

simplify a bit

Answer 8

cant we sub sin theta? :/

Answer 9

you can do anything you want to :)

Answer 10

do it by sin theta,dk how to proceed

Answer 11

what is the problem with the above method? :|

Answer 12

nothing but do with sin theta :|

Answer 13

if u use arcsinx=t say x changes to sint and dx changes cost dt

Answer 14

im saying x^2=sint

Answer 15

so denominator->cost so tantsect integrates to sect

Answer 16

x=sint is what you mean

Answer 17

yes sorryyyyyyyyyyyyyyy

Answer 18

so..
sect is what we get after integrating it..
and t=arcsin(x)
so..
sec(arcsin(x))
\[\sec(\sec^{-1}(\frac{1}{1-x^2}))=\frac{1}{1-x^2}\]

Answer 19

but do we have to substitute sin t in whole equation or just this? :| (I2) say

Answer 20

I2(say what)

Answer 21

\[\Large \int\limits\limits_{}^{}\frac{x^2}{\sqrt{1-x^2}}=I2\]

Answer 22

hmm..

Answer 23

o.O