Question

@ganeshie8 :)

Answer 1

@iambatman

Answer 2

HELP?:)

Answer 3

2 arcsin x +C

Answer 4

thats what i got:)

Answer 5

2 arc sin sqrtx +C

Answer 6

@ganeshie8 is it right?

Answer 7

im not sure, i would complete the square for the denominator first

Answer 8

I think you can by pass the completing the square if you let u=sqrt(x)
then u^2=x
so 2u du=dx

Answer 9

\[\int\limits_{}^{}\frac{ 2 u du}{u \sqrt{1-u^2}}\]

Answer 10

so whats my final answer?

Answer 11

Oh nice !

Answer 12

its that?

Answer 13

recall that
\[\int\limits_{}^{}\frac{1}{\sqrt{1-x^2}} dx=\arcsin(x)+C\]

Answer 14

Our u's up there canceled

Answer 15

and u was sqrt(x)

Answer 16

your answer is right

Answer 17

this is correct?

Answer 18

2 arc sin sqrtx +C

this second one you mentioned anyways

Answer 19

do you need to put +C

Answer 20

or does it already put it on the outside of the box

Answer 21

yes its included in the problem alreay:)

Answer 22

outside of the box:)

Answer 23

looks it is not arpreting `arc` correctly.... notice that it is slanty as regular text

Answer 24

*interpreting

Answer 25

so you think he should enter it in as
\[\sin^{-1}(\sqrt{x})+C\]

Answer 26

2 times that

Answer 27

I don't know what font styles to look for

Answer 28

he got it right it seems since he is silent :) @liz_mustang

Answer 29

i guess math isn't in italic

Answer 30

or shouldn't be

Answer 31

It's strange, I got a different answer when I completed the square: sin^-1(2x -1) + C

Answer 32

answers can look different in integration

Answer 33

but that doesn't mean they are different

Answer 34

I see, do you have a reference that explains that in more detail? Wolfram got an answer that's really different.

Answer 35

lol im refering to latex... regular text is italic but the commands understood by latex compiler are not italic
latex command :
\[\arcsin(x)\]
latex regular text :
\[arcsin(x) \]

Answer 36

\[\int \dfrac{1}{\sqrt{x}\sqrt{1-x}} dx = \int
\dfrac{1}{\sqrt{x-x^2}} dx = \int
\dfrac{2}{\sqrt{1-(2x-1)^2}} dx \]

Answer 37

\[\int\limits_{}^{}\frac{1}{\sqrt{x-x^2}}dx=\int\limits_{}^{}\frac{ 1}{\sqrt{-x^2+x+(\frac{1}{2})^2-(\frac{1}{2})^2}} dx \\ =\int\limits_{}^{} \frac{1}{\sqrt{-(x-\frac{1}{2})^2+\frac{1}{4}}} dx =\int\limits_{}^{} \frac{1}{\sqrt{\frac{1}{4}}} \frac{1}{\sqrt{-4(x-\frac{1}{2})^2+1}} dx \\ =\int\limits_{}^{}2 \frac{1}{\sqrt{-(2(x-\frac{1}{2}))^2+1}}dx =2\int\limits_{}^{} \frac{1}{\sqrt{-(2x-1)^2+1}} dx \\ \text{ \let } u=2x-1 \\ du=2 dx \\ \int\limits_{}^{} \frac{du}{\sqrt{1-u^2}} =\arcsin(u)+C =\arcsin(2x-1)+C\]arcsin(2x-1) and 2arcsin(sqrt(x)) should just differ by a constant
off the top of my head I don't know what constant that is ...

Answer 38

Oh i see all these different answers might be related to the integration constnat

Answer 39

\[\arcsin(2x-1)=2 \arcsin (\sqrt{x}) +C \\ 2x-1=\sin(2 \arcsin(\sqrt{x})+C ) \\ 2x-1 =\sin(2 \arcsin(\sqrt{x}))\cos(C)+\cos(2 \arcsin(\sqrt{x}))\sin(C) \] \[2x-1= \\ 2 \sin(\arcsin(\sqrt{x}))\cos(\arcsin(\sqrt{x}))\cos(C)+(\cos^2(\arcsin(\sqrt{x}))-\sin^2(\arcsin(\sqrt{x})))\sin(C) \\ \]

\[2x-1=2 \sqrt{x} \sqrt{1-x}\cos(C)+(1-x-x)\sin(C) \\ 2x-1 =2 \sqrt{x}\sqrt{1-x}\cos(C)+\sin(C) \\\]
Trying to see if I can find what constant a differ by

Answer 40

that one line got a little cut off

Answer 41

oops I made an error

Answer 42

\[2x-1=2 \sqrt{x} \sqrt{1-x}\cos(C)+(1-2x)\sin(C)\]

Answer 43

ahh i forgot almost all these inverse trig identities
http://www.wolframalpha.com/input/?i=%282arcsin%28sqrt%28x%29%29+-+arcsin%282x-1%29%29%27

Answer 44

I think we want sin(C)=-1 and cos(C)=0
so C=3pi/2

Answer 45

\[\arcsin(2x-1)=2\arcsin(\sqrt{x})+\frac{3\pi}{2}\]

Answer 46

or I guess we could have said C was 3pi/2+2npi

Answer 47

wait maybe it shouldn't be 3pi/2
since arcsin is a number between -pi/2 and pi/2

Answer 48

-pi/2?

Answer 49

\[\arcsin(2x-1)=2\arcsin(\sqrt{x})-\frac{\pi}{2}\]

Answer 50

lol I don't know
but I think it should be a number C such that sin(C)=-1

Answer 51

brilliant !!!! its working perfectly, when x = 0 : arcsin(-1) = -pi/2 as expected etc etc..

Answer 52

x=1/2
arcsin(2*1/2-1)=arcsin(0)=0
2arcsin(sqrt(1/2))-pi/2=2*pi/4-pi/2=pi/2-pi/2=0

Answer 53

i guess it does seem to work

Answer 54

I wish wolfram would have confirmed it as an identity though

Answer 55

it is an identity, graph confirms
http://gyazo.com/ce41ef8b8ea3932c6afb1db8736c0185

Answer 56

neatness