how to (1 Viewer)

[Buiboi]

integrate inverse sinx?

 

[SoulSearcher]

I take it you don't know integration by parts? Well, if I remember correctly, the indefinite integral of sin^-1x is xsin^-1x + sqrt(1-x²) + constant, usually they would give a sort of a lead in question to help you find the integral of arcsin x, since I dont think that it is possible to integrate it directly using extension 1 methods.

 

[the-1-n-only-me]

last time i heard 3u people didnt have to integrate inverse trig.

but if you're talking about area under a inverse trig curve. just convert it normal trig and change the limits.

 

[JasonNg1025]

yeah integration by parts should work

let's try

(I use arcsin(x) for inverse sine)

I = int(arcsin(x)dx)

u = arcsin(x), dv = dx.
du = 1/(sqrt(1-x^2)) (dx), v = x

I = xarcsin(x) - int(x/(sqrt(1-x^2)))

use substitution u = 1-x^2

if you multiply the integral by -1/2 you get a -2x up the top

and you just get -int(1/sqrt(u)du) which should be easy enough

hope it's not too hard to read

(this is if you havn't managed to do it yet )

and I thought integration by parts was in the extension 2 course... so if you just do extension 1 I'd leave it

and oh wow you actually remembered it?

 

[SoulSearcher]

last time i heard 3u people didnt have to integrate inverse trig.

but if you're talking about area under a inverse trig curve. just convert it normal trig and change the limits.

If that was the case, it'd be a lot easier to do that than memorise the integral of arcsin x.

and oh wow you actually remembered it?

After doing the question a few times it just stuck in my head.

 

[Chinmoku03]

IIRC, an inverse sine curve is just a sine curve about the y axis, so you can use the V = pi x Sx²dy, where S is the integration symbol, cos I dunno where to find one >.>;

 

[Buiboi]

IIRC, an inverse sine curve is just a sine curve about the y axis, so you can use the V = pi x Sx²dy, where S is the integration symbol, cos I dunno where to find one >.>;

OOOOHHHHHH YEAH!!...i remember using this way to do it b4...

CHEERS!