[smith50]

How would you intergrate ;
I= Sinx(raised to the negative 1 )???
Thanks in advance

[Intriguing Alias]

(Original post by smith50)
How would you intergrate ;
I= Sinx(raised to the negative 1 )???
Thanks in advance

Right now is it OR

[Moiraclaire]

(sinx)^-1 is otherwise known as cosecx ? then there is a formula book...

or sin^{-1}(x) is FP3, so I'm guessing that's not the question? If it is I can try and explain it at a C4 level, you would need to use by parts with u = arcsinx and dv/dx = 1

[smith50]

(Original post by hassi94)
Right now is it OR

The first one sinx-1

[Intriguing Alias]

(Original post by smith50)
The first one sinx-1

Seems outside of core 4's jurisdiction But you'd integrate by parts and then by substitution/inspection.

[smith50]

(Original post by hassi94)
Seems outside of core 4's jurisdiction But you'd integrate by parts and then by substitution/inspection.

Did so but got confused

[Extricated]

Pretty sure you've misread the question, that isn't c4

[PhysicsGirl]

Integrating arcsin(x) in AQA C4 is usually done like this, I think;
let y=arcsin(x) therefore x= sin(y)
therefore dx/dy = cos(y)
you want dy/dx, so you inverse that, and get dy/dx= 1/ cos(y)
rearranging the identity sin^2(x)+cos^2(x)=1, cos(y) = sqrt [ 1- sin^2(y) ]
therefore, substituting that into dy/dx, dy/dx= 1/ sqrt[ 1-sin^2(y) ]
and as stated originally, sin(y) = x, so sin^2(y) = x^2
therefore, dy/dx = 1/ sqrt [1 - x^2].
Hope that helps

[PhysicsGirl]

By the way, if you are on AQA C4, this is given in your formula book for you, so you don't need to worry too much about how to derive it!

[Moiraclaire]

(Original post by smith50)
The first one sinx-1

Hassi's first one and sinx-1 seem totally different !

it will be (sinx)^-1 which is cosecx.

I think exam questions refrain from putting things in the way Hassi's first one is written (I don't know how to write in maths on here) as it can be interpreted as an inverse function. (arcsinx)

[smith50]

(Original post by Extricated)
Pretty sure you've misread the question, that isn't c4

It's an old spec question.