# Integral of Arcsin[x]

by Jameson
Tags: arcsinx, integral
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 P: 787 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.
 Emeritus Sci Advisor PF Gold P: 16,091 If only you were working with its derivative! I bet you know how to integrate something that looks like its derivative!
 P: 787 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.
 P: 666 Integral of Arcsin[x] Use integration by parts. Remember to let u=arcsin(x) and v=x. u sub: InverseLogAlgebraicTrigExp
 Emeritus Sci Advisor PF Gold P: 16,091 Sheesh, just give him the answer, why don't ya?
P: 787
 Quote by apmcavoy 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.
 P: 666 I apologize Hurkyl
 P: 84 well, you know the integral of sinx with limits. Now arcsin x will be the limits, and you can make a rectangle.
 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.
 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
 Sci Advisor HW Helper P: 3,144 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 $$\int \sin^{-1} x dx = \sqrt {1-x^2} + x \sin^{-1} x + C$$
 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} 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.
 HW Helper P: 3,348 For the second time you integrate by parts, swap your choices.
 P: 17 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 :(
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 Quote by 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.
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.
 Emeritus Sci Advisor PF Gold P: 4,500 After the first integration by parts, I would use a substitution
 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
 P: 1 use seperation 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

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