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Integral of Arcsin[x] 
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#1
Sep1505, 06:44 PM

P: 788

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 [tex]\int \arcsin{(x)}dx = \sqrt{1x^2}+ x\arcsin{(x}) + C[/tex] from my calculator and mathematica. Hint please. 


#2
Sep1505, 06:47 PM

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



#3
Sep1505, 06:55 PM

P: 788

Sure... I'd draw a nice triangle and do trig substitution if it was [tex]\int \frac{dx}{\sqrt{1x^2}}[/tex]
Hmmm...I'll think out loud here. If the integral has a squareroot and is in the form of [itex]c^2x^2[/itex] 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 [itex]\sqrt{a^2+b^2}[/itex], it seems that one leg might need to be [itex]\sqrt{\arcsin{(x)}}[/itex] Somehow I don't think I'm on track. 


#4
Sep1505, 07:01 PM

P: 668

Integral of Arcsin[x]
Use integration by parts. Remember to let u=arcsin(x) and v=x.
u sub: InverseLogAlgebraicTrigExp 


#5
Sep1505, 07:02 PM

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



#6
Sep1505, 07:06 PM

P: 788

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


#7
Sep1605, 12:05 PM

P: 668

I apologize Hurkyl



#8
Sep1605, 06:39 PM

P: 84

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



#9
Sep1605, 08:00 PM

P: n/a

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



#10
Sep1705, 10:51 AM

P: 3

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



#11
Sep1905, 11:07 PM

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Just for the fun of it ...
The sum of the integrals [tex]\int \sin^{1} x dx + \int \sin y dy[/tex] is just the area of the bounding rectangle: [itex] x \times \sin^{1} x[/itex] Since [tex]\int \sin y dy = \cos y + C[/tex] and [tex]\cos y = \cos \sin^{1} x = \sqrt {1x^2}[/tex] it follows that [tex]\int \sin^{1} x dx = \sqrt {1x^2} + x \sin^{1} x + C[/tex] 


#12
Jun907, 05:00 AM

P: 17

Hey, Im 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} xarcsinxS(x*d(arcsinx))= xarcsinxS(x/(1x^2)^0.5= {u(x)=x v'(x)=arcsinx} xarcsinxxarcsinx+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. 


#13
Jun907, 05:18 AM

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



#14
Jun907, 05:51 AM

P: 17

Im sorry, Tried it also and all i got is:
xarcsine x (x^2/2)(1/(Sqrt(1x^2)))(x/4)(sqrt(1/(1x^2))+1/4(arcsinx) this isn't going anywhere :( 


#15
Jun907, 02:03 PM

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#16
Jun907, 04:37 PM

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



#17
Jun907, 04:43 PM

P: 17

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 


#18
Oct1008, 01:03 PM

P: 1

use seperation by parts
u=arcsinx du=dx/(1x^2)^1/2 dv=dx v=x uvingegral vdu = xarsinxintegral x/(1x^2)^1/2 use u substitution with u = 1x^2 so du = 2x than you get xarsinx + (1x^2)^1/2 + c 


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