Integral problemWatch

#1
Hi, I am stuck on an integral:
I = Integral of sqrt(1-x^2)

at first I thought, easy - i'll just use a trig substitution (x = sin[u]..) and solved it, ended up with a bit of a messy solution of :
1/4 (sin[2*arcsin[x]] + 2*arcsin[x])

then thought, why did i choose u = sin[x]? i'll do it with x = cos[u] to ensure I get the same result, I got:
1/4 (sin[2*arccos[x]] - 2*arccos[x])

Can anyone think of reasons why I got different solutions? I thought may be they are an identity? but I couldn't prove if they were so i'm still not sure if theyre right

Any ideas or solutions will be great

p.s keeping this to real plane, i.e |x| =< 1, so the substitution x = sinu or = cosu should be fine
also, + constants should be there but I'm lazy :P
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6 years ago
#2
Isn't it an indefinite integral? It should be.
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6 years ago
#3
Sorry, I do not have time today to give a full solution or to check your answers carefully. However, I will give you a brief justification and maybe get chance to provide more substance later.

It is down to the constant of integration (or + c) which you chose not to include in your solutions

arcsinx and arccosx only differ by a constant. Ultimately, your solutions are the same with a different constant of integration.
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6 years ago
#4
U probably didnt do it right. Let x=sinu, dx=cosudu. Your integral transforms to cos^2u whicb is easily integral if u say cos^2u= (cos2u -1)/2.
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#5
(Original post by mathslearn)
arcsinx and arccosx only differ by a constant. Ultimately, your solutions are the same with a different constant of integration.

(Original post by iluvmaths)
U probably didnt do it right. Let x=sinu, dx=cosudu. Your integral transforms to cos^2u whicb is easily integral if u say cos^2u= (cos2u -1)/2.
My integrals are correct :P if you follow what you said through you get,
I = 1/4 (sin2u + 2u) then substitute back, u = arcsin[x], and you get one of the answers. Thank you still
btw I assume you meant cos^2[u] = (cos[2u] + 1)/2
Last edited by pi+e=pie; 6 years ago
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6 years ago
#6
Part (b) of the ziedj problem sheet:

To ignore the constant of integration, introduce limits and and show that the two solutions are equal under evaluation.

Hint: You may find it useful to use the identity

[4 marks]
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6 years ago
#7
Hahaha. PRSOM.

OP:
You should know already that an indefinite integral gives a family of solutions, since antiderivatives are non-unique inverses of derivatives.
This is understanding which is required at the level you are, and I consider only ziedj's comment relevant.
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6 years ago
#8
(Original post by pi+e=pie)
Hi, I am stuck on an integral:
I = Integral of sqrt(1-x^2)

at first I thought, easy - i'll just use a trig substitution (x = sin[u]..) and solved it, ended up with a bit of a messy solution of :
1/4 (sin[2*arcsin[x]] + 2*arcsin[x])

then thought, why did i choose u = sin[x]? i'll do it with x = cos[u] to ensure I get the same result, I got:
1/4 (sin[2*arccos[x]] - 2*arccos[x])

Can anyone think of reasons why I got different solutions? I thought may be they are an identity? but I couldn't prove if they were so i'm still not sure if theyre right

Any ideas or solutions will be great

p.s keeping this to real plane, i.e |x| =< 1, so the substitution x = sinu or = cosu should be fine
also, + constants should be there but I'm lazy :P
The two solution are the same. THe diference between them only one constant
For the 1st part
1/4sin[2*arccos(x)]=1/4*2*sin(arccosx)*cos(arccosx)=1/2x*sqrt(1-x^2)
at the first solution
1/4sin(2*arcsinx)=1/4*2*sin(arcsinx)*cos(arcsinx)=1/2x*sqrt(1-x^2)
For the second part

So this part differ only in a constant
with denote of C will be an integration constant just as at another solution
At definite integration these constants will be cancelled out
Last edited by ztibor; 6 years ago
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6 years ago
#9
could someone help me out with this question please given than sin (x+y) = 3cos(x-Y), express tan y in terms of tan x
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6 years ago
#10
(Original post by mskh1)
could someone help me out with this question please given than sin (x+y) = 3cos(x-Y), express tan y in terms of tan x
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6 years ago
#11
(Original post by gff)
Isn't it an indefinite integral? It should be.
(Original post by gff)
Hahaha. PRSOM.

OP:
You should know already that an indefinite integral gives a family of solutions, since antiderivatives are non-unique inverses of derivatives.
This is understanding which is required at the level you are, and I consider only ziedj's comment relevant.
Would neg you again though on recharge.

OP, this is easily / easier seen by remembering that cosine and sine are really just the same function with a different translation each time.

So cos(x) = sin(x + pi/2) [is this correct? Roughly two years since having to do trig identities ... ] so you are doing two substitution (technically) with the latter just a linear substitution.
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6 years ago
#12
(Original post by Oh I Really Don't Care)
Would neg you again though on recharge.

OP, this is easily / easier seen by remembering that cosine and sine are really just the same function with a different translation each time.

So cos(x) = sin(x + pi/2) [is this correct? Roughly two years since having to do trig identities ... ] so you are doing two substitution (technically) with the latter just a linear substitution.
In case you have any constructive feedback, then fine, share it.
Otherwise, don't feel responsible to inform me next time when you want to neg me "again".
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6 years ago
#13
(Original post by gff)
In case you have any constructive feedback, then fine, share it.
Otherwise, don't feel responsible to inform me next time when you want to neg me "again".
It was obvious I had constructive feedback considering I was the only person to answer the question correctly - my issue was the fact you have absolutely no clue what you are talking about and even then have the nerve to say things such as "at this level it is expected you know this" when you yourself have virtually zero knowledge of mathematics.
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6 years ago
#14
I have the feeling that you are the one who has too much nerves to spend.
Anyway, this not my problem, and I recommend that you don't make it mine.

How sure are you that I have absolutely no clue what I am talking about? How did you estimate anything about me?
I'm studying at the same level as the OP, am doing the same problems, and I'm expected to have an idea about all this.

Perhaps, when you were at my level, you had more knowledge of Mathematics than me.
Unfortunately, we can't all be the same, can we?
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6 years ago
#15
(Original post by gff)
I have the feeling that you are the one who has too much nerves to spend.
Anyway, this not my problem, and I recommend that you don't make it mine.

How sure are you that I have absolutely no clue what I am talking about? How did you estimate anything about me?
I'm studying at the same level as the OP, am doing the same problems, and I'm expected to have an idea about all this.

Perhaps, when you were at my level, you had more knowledge of Mathematics than me.
Unfortunately, we can't all be the same, can we?
I am not really concerned with your level, I come here to assist others who require help. When you not only offer poor tips, but to knock someones confidence (who is attempting the problem marked at university level, and I think you are a high school student so no, not the same levels ...) it is hardly surprising you will be called up on this.

Please carry on posting here, I rarely do anymore and many of the other mathmos at uni don't when it is term time also, so we do need new posters - however, please try to offer meaningful hints or suggestions, not vague musings of no worth.
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6 years ago
#16
(Original post by Oh I Really Don't Care)
...
That was the kind of feedback I was looking for.
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