What is the integral of Arcsin[x]?

[Jameson]

I just can't see it. I would think that it would involve some form of trig substitution, but I'm just drawing a blank. I'll do the work if someone can please give me a nice little hint.

I know that \int \arcsin{(x)}dx = \sqrt{1-x^2}+ x\arcsin{(x}) + C from my calculator and mathematica.

Hint please.

 

[Hurkyl]

If only you were working with its derivative! I bet you know how to integrate something that looks like its derivative!

 

[Jameson]

Sure... I'd draw a nice triangle and do trig substitution if it was \int \frac{dx}{\sqrt{1-x^2}}

Hmmm...I'll think out loud here. If the integral has a square-root and is in the form of c^2-x^2 then x is one leg, c is the hypotenuse, and the other leg is the previously mentioned radical.

Is trig substitution with right triangles on the right track? Since the hypotenuse is \sqrt{a^2+b^2}, it seems that one leg might need to be \sqrt{\arcsin{(x)}} Somehow I don't think I'm on track.

 

[amcavoy]

Use integration by parts. Remember to let u=arcsin(x) and v=x.

u sub: InverseLogAlgebraicTrigExp

 

[Hurkyl]

Sheesh, just give him the answer, why don't ya?

 

[Jameson]

Use integration by parts. Remember to let u=arcsin(x) and v=x.

u sub: InverseLogAlgebraicTrigExp

Ah, thank you! It's so simple.

And thank you Hurkyl as well. I still have lots to learn.

 

[amcavoy]

I apologize Hurkyl

 

[tongos]

well, you know the integral of sinx with limits. Now arcsin x will be the limits, and you can make a rectangle.

 

[MalleusScientiarum]

Or you could just take the derivative of the right hand side and go "ta da!" and that's proof enough for me.

 

[professorlucky]

hint

equate the arc sine to another variable e.g y.making it a sine fxn.e.g the arc sine of 0.5=30,while sine30=0.5.this will simplify the integral and further substitution will conclude it

 

[Tide]

Just for the fun of it ...

The sum of the integrals

\int \sin^{-1} x dx + \int \sin y dy

is just the area of the bounding rectangle: x \times \sin^{-1} x

Since \int \sin y dy = -\cos y + C and \cos y = \cos \sin^{-1} x = \sqrt {1-x^2} it follows that
\int \sin^{-1} x dx = \sqrt {1-x^2} + x \sin^{-1} x + C

 

[FlashStorm]

Integration by parts?

Hey, I am really sorry to arise dead threads from the past (which i have seen though google) but somehting really weird happened me when I tried to use integration by parts on arcsinx.


let me show you:
S(arcsinx)= {v'(x)=1} {u(x)=arcsinx}

xarcsinx-S(x*d(arcsinx))=

xarcsinx-S(x/(1-x^2)^0.5= {u(x)=x v'(x)=arcsinx}

xarcsinx-xarcsinx+S(arcsin)dx


==> S(arcsinx)=S(arcsinx)

:\
I know I have done something really stupid here, but please be easy on me since I started studying Integrals only three days ago.

In the second time I used integration by parts, do I miss something , is there another efficient choice of v and u?

Thanks in advance,
Aviv

p.s: I will edit it better to use normal math signs once I figure out how.

 

[Gib Z]

For the second time you integrate by parts, swap your choices.

 

[FlashStorm]

Im sorry, Tried it also and all i got is:

xarcsine x -(x^2/2)(1/(Sqrt(1-x^2)))-(x/4)(sqrt(1/(1-x^2))+1/4(arcsinx)

this isn't going anywhere :(

 

[dextercioby]

I just can't see it. I would think that it would involve some form of trig substitution, but I'm just drawing a blank. I'll do the work if someone can please give me a nice little hint.

I know that \int \arcsin{(x)}dx = \sqrt{1-x^2}+ x\arcsin{(x}) + C from my calculator and mathematica.

Hint please.

Make x=\sin t. Then apply part integration on the resulting integral. It's just a way to avoid the simple solution of part integrating directly.

 

[Office_Shredder]

After the first integration by parts, I would use a substitution

 

[FlashStorm]

solve it officially. Fixed during some major mistakes I had about dev' and stuff.

did it without subtition, only using integration by parts.

if you are interested what I did then you are welcome to tell me to write my solution.Thanks guys :)
gg

 

[josiahsavino]

use separation by parts

u=arcsinx
du=dx/(1-x^2)^1/2

dv=dx
v=x

uv-ingegral vdu = xarsinx-integral x/(1-x^2)^1/2
use u substitution with u = 1-x^2 so du = -2x
than you get

xarsinx + (1-x^2)^1/2 + c

 

[Liqorice]

f(x)=arcsinx f'(x)=1/radical(1-xsquare)
g'(x)=1 g(x)=x

// S means integral

S arcsin x dx= x arcsin x - S xdx/radical(1-xsquare) = x arcsin x + S (radical(1-xsquare))'dx=
=x arcsin x + radical(1-xsquare) +C