Is this question possible? (1 Viewer)

[narrowpin]

how do u find the primitive of this: (1 - x^2)^1/2

 

[uniform]

It is possible, but have you learned integration techniques such as substitution yet?

..because you can find the primitive by letting x=sin@ or x=cos@

 

[narrowpin]

no we havent....but can you just solve it for me please. Thanks in advance=)

 

[watatank]

how do u find the primitive of this: (1 - x^2)^1/2

so you want to find prim (1 - x²)^0.5

let x = sin@

so now it becomes

prim (1-sin²@)^0.5

= prim (cos²@])^0.5

= prim (cos@)

= sin@ + c

= x + c

 

[YannY]

lol you're funny watatank. Try to differetiate your solution

 

[Slidey]

how do u find the primitive of this: (1 - x^2)^1/2

In the 3u exam, it is assumed that you will be given all substitutions to carry out. That said, I highly recommend you learn how to integrate without being given the substitutions, as it increases your speed and accuracy.

 

[Slidey]

so you want to find prim (1 - x²)^0.5

let x = sin@

so now it becomes

prim (1-sin²@)^0.5

= prim (cos²@])^0.5

= prim (cos@)

= sin@ + c

= x + c

Careful. This is what happens when you forget the differential.

Let
x=sin@
dx=cos@ d@

Int (1-sin^2(@))^1/2 cos@ d@
= Int cos^2(@) d@

Use half-angle formula (double angle formula essentially):
cos^2(x) = (1+cos(2x))/2

Int cos^2(@) d@
= Int (1+cos(2@))/2 d@
= @/2+sin(2@))/4 + c
arcsin(x)=@
So the answer is: arcsin(x)/2 + sin(2arcsin(x))/4 + c

Wow. That looks funky. It's been three years since I've done trigonometric substitution, so correct me if I'm wrong.

 

[YannY]

Careful. This is what happens when you forget the differential.

Let
x=sin@
dx=cos@ d@

Int (1-sin^2(@))^1/2 cos@ d@
= Int cos^2(@) d@

Use half-angle formula (double angle formula essentially):
cos^2(x) = (1+cos(2x))/2

Int cos^2(@) d@
= Int (1+cos(2@))/2 d@
= @/2+sin(2@))/4 + c
arcsin(x)=@
So the answer is: arcsin(x)/2 + sin(2arcsin(x))/4 + c

Wow. That looks funky. It's been three years since I've done trigonometric substitution, so correct me if I'm wrong.

Thats correct.

 

[uniform]

Let
x=sin@
dx=cos@ d@

Int (1-sin^2(@))^1/2 cos@ d@
= Int cos^2(@) d@

Use half-angle formula (double angle formula essentially):
cos^2(x) = (1+cos(2x))/2

Int cos^2(@) d@
= Int (1+cos(2@))/2 d@
= @/2+sin(2@))/4 + c
arcsin(x)=@
So the answer is: arcsin(x)/2 + sin(2arcsin(x))/4 + c

Wow. That looks funky. It's been three years since I've done trigonometric substitution, so correct me if I'm wrong.

Yep, that should be correct,
maybe it would look less funky if you simplified sin(2arcsin(x))/4 to [xsqrt(1-x^2)]/2 to get
(1/2)[arcsin(x) + xsqrt(1-x^2)]

 

[Slidey]

I'll torture myself by deriving it. Le sigh:

arcsin(x)/2 + sin(2arcsin(x))/4 + c
= arcsin(x)/2 + sin(arcsinx).cos(arcsinx)/2 + c
= (arcsinx + x.cos(arcsinx) + k)/2
= (arcsinx + x.sqrt(1-x^2) + k)/2

Derivative is: (1/sqrt(1-x^2) + sqrt(1-x^2) - x^2/sqrt(1-x^2))/2
= (1/sqrt(1-x^2) + (1-x^2)/sqrt(1-x^2) - x^2/sqrt(1-x^2))/2
= (1+1-x^2-x^2)/(2.sqrt(1-x^2))
= (1-x^2)/sqrt(1-x^2)
= sqrt(1-x^2)

Awesome!

 

[YannY]

Yep, that should be correct,
maybe it would look less funky if you simplified sin(2arcsin(x))/4 to [xsqrt(1-x^2)]/2 to get
(1/2)[arcsin(x) + xsqrt(1-x^2)]

Hey how do get sin(2arcsin(x))/4 to [xsqrt(1-x^2)]/2 . did you use pythagoras?

 

[Slidey]

Use double angle formulae:

sin(2arcsin(x))=2sin(arcsin(x))cos(arcsin(x))=2xcos(arcsin(x))

From there, you can draw a triangle and use soh/cah/toa or you can use trig identities:

cos^2(x)+sin^2(x)=1
cosx=sqrt(1-sin^2(x))
cos(arcsin(x))=sqrt(1-sin^2(arcsinx))
cos(arcsin(x))=sqrt(1-x^2)

EDIT: The key thing to remember is that arcsin(x) is an ANGLE. As is any inverse trig function.

 

[S1M0]

Hey how do get sin(2arcsin(x))/4 to [xsqrt(1-x^2)]/2 . did you use pythagoras?

LOL!

 

[YannY]

LOL!

Yeah i was stupid. Right you are to humiliate me.

Thanks slidey and do we have to know that for 3u? and what other inverse trig identities do we need to know?

 

[Slidey]

Yeah i was stupid. Right you are to humiliate me.

Thanks slidey and do we have to know that for 3u? and what other inverse trig identities do we need to know?

Knowing how to integrate sin^2(x) is in the syllabus.

I don't know if knowing how to simplify arcsin(cos(x)) is, but at a guess, I'd say yes. I do recall textbooks having sections on questions like that.

How do you figure them out? Draw a right-angle triangle. suppose it's sin(arctan(x)):

You've got x right? That's x/1. So opposite over adjacent, x is opposite the angle arctan(x), and the adjacent side has length one. Hypotenuse is 1+x^2. Sin is opposite over hypotenuse, so arcsin is arcsin(x/(1+x^2)).

Thus sin(arctan(x)) = sin(arcsin(x/(1+x^2)) = x/(1+x^2).

What is the value of cos(arctan(x/(x+1)))?

LOL!

LOL what? It's not like it's straightforward or well known.

 

[watatank]

Careful. This is what happens when you forget the differential.

Let
x=sin@
dx=cos@ d@

Int (1-sin^2(@))^1/2 cos@ d@
= Int cos^2(@) d@

Use half-angle formula (double angle formula essentially):
cos^2(x) = (1+cos(2x))/2

Int cos^2(@) d@
= Int (1+cos(2@))/2 d@
= @/2+sin(2@))/4 + c
arcsin(x)=@
So the answer is: arcsin(x)/2 + sin(2arcsin(x))/4 + c

Wow. That looks funky. It's been three years since I've done trigonometric substitution, so correct me if I'm wrong.

lol. yeah my bad. i thought that answer looked too neat

 

[narrowpin]

i think this might be a dumb question....but whats arc??

Thank for all the help though!

 

[watatank]

arcsin is the inverse sine function.

 

[A High Way Man]

inverse functions are defined when the function is one-to-one (i forget the fancy word).. when stating an inverse function remember to state its domain and range.. ie [-Pi/2, Pi/2].. otherwise, its undefined and the universe will explode.

 

[narrowpin]

i get it! thank you all for the help