## Integral of Arcsin[x]

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.

 Recognitions: Gold Member Science Advisor Staff Emeritus If only you were working with its derivative! I bet you know how to integrate something that looks like its derivative!
 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.

## Integral of Arcsin[x]

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

u sub: InverseLogAlgebraicTrigExp
 Recognitions: Gold Member Science Advisor Staff Emeritus Sheesh, just give him the answer, why don't ya?

 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.
 I apologize Hurkyl
 well, you know the integral of sinx with limits. Now arcsin x will be the limits, and you can make a rectangle.
 Or you could just take the derivative of the right hand side and go "ta da!" and that's proof enough for me.
 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
 Recognitions: Homework Help Science Advisor 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 $$\int \sin^{-1} x dx = \sqrt {1-x^2} + x \sin^{-1} x + C$$
 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.
 Recognitions: Homework Help For the second time you integrate by parts, swap your choices.
 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.