integral of $int frac{du}{sqrt{(1-u^2)}}$
1
$begingroup$
I'm curious how my textbook got:
$$int frac{du}{sqrt{(1-u^2)}}$$
to
$$sin^{-1}(u)+ c$$
This considering that "$u$" could be simplified into $u = sin(x)$
Thus we get:
$$int frac{du}{sqrt{(1-sin^2(u))}}$$
Which is further simplified into:
$$int frac{du}{sqrt{(cos^2(u))}}$$
Since: $$cos^2(u) =sqrt{(1-sin^2(u)}$$
Meaning that we get:
$$int frac{du}{cos(u)}$$
Which is not $sin^{-1}(u)+ c$ when integrated.
What am I doing wrong?
calculus integration
edited Dec 30 '18 at 0:33
Larry
2,55531131
asked Dec 30 '18 at 0:22
oxodooxodo
240110
$endgroup$
add a comment |
1
$begingroup$
I'm curious how my textbook got:
$$int frac{du}{sqrt{(1-u^2)}}$$
to
$$sin^{-1}(u)+ c$$
This considering that "$u$" could be simplified into $u = sin(x)$
Thus we get:
$$int frac{du}{sqrt{(1-sin^2(u))}}$$
Which is further simplified into:
$$int frac{du}{sqrt{(cos^2(u))}}$$
Since: $$cos^2(u) =sqrt{(1-sin^2(u)}$$
Meaning that we get:
$$int frac{du}{cos(u)}$$
Which is not $sin^{-1}(u)+ c$ when integrated.
What am I doing wrong?
calculus integration
edited Dec 30 '18 at 0:33
Larry
2,55531131
asked Dec 30 '18 at 0:22
oxodooxodo
240110
$endgroup$
add a comment |
1
1
1
$begingroup$
I'm curious how my textbook got:
$$int frac{du}{sqrt{(1-u^2)}}$$
to
$$sin^{-1}(u)+ c$$
This considering that "$u$" could be simplified into $u = sin(x)$
Thus we get:
$$int frac{du}{sqrt{(1-sin^2(u))}}$$
Which is further simplified into:
$$int frac{du}{sqrt{(cos^2(u))}}$$
Since: $$cos^2(u) =sqrt{(1-sin^2(u)}$$
Meaning that we get:
$$int frac{du}{cos(u)}$$
Which is not $sin^{-1}(u)+ c$ when integrated.
What am I doing wrong?
calculus integration
edited Dec 30 '18 at 0:33
Larry
2,55531131
asked Dec 30 '18 at 0:22
oxodooxodo
240110
$endgroup$
I'm curious how my textbook got:
$$int frac{du}{sqrt{(1-u^2)}}$$
to
$$sin^{-1}(u)+ c$$
This considering that "$u$" could be simplified into $u = sin(x)$
Thus we get:
$$int frac{du}{sqrt{(1-sin^2(u))}}$$
Which is further simplified into:
$$int frac{du}{sqrt{(cos^2(u))}}$$
Since: $$cos^2(u) =sqrt{(1-sin^2(u)}$$
Meaning that we get:
$$int frac{du}{cos(u)}$$
Which is not $sin^{-1}(u)+ c$ when integrated.
What am I doing wrong?
calculus integration
calculus integration
edited Dec 30 '18 at 0:33
Larry
2,55531131
asked Dec 30 '18 at 0:22
oxodooxodo
240110
edited Dec 30 '18 at 0:33
Larry
2,55531131
asked Dec 30 '18 at 0:22
oxodooxodo
240110
edited Dec 30 '18 at 0:33
Larry
2,55531131
edited Dec 30 '18 at 0:33
Larry
2,55531131
edited Dec 30 '18 at 0:33
Larry
2,55531131
2,55531131
asked Dec 30 '18 at 0:22
oxodooxodo
240110
asked Dec 30 '18 at 0:22
oxodooxodo
240110
asked Dec 30 '18 at 0:22
oxodooxodo
240110
240110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
3
$begingroup$
So, when you make the substitution $u=sin x$, you also need to make the substitution for $du$ as well:
$$du=frac{du}{dx}dx=cos xdx$$
Thus, the new integral becomes:
$$int frac{du}{sqrt{1-u^2}}=int frac{dxcos x}{sqrt{1-sin^2x}}$$
Hopefully, you can take it from there.
answered Dec 30 '18 at 0:26
Noble MushtakNoble Mushtak
15.4k1835
$endgroup$
$begingroup$
I do get that. I should've maybe added that the original integral was : $$int frac{du}{sqrt{(1-(x-1)^2)}}$$ . The author then substitued $(x-1)$ with $u$ giving me $du = dx$. We then get $$int frac{du}{sqrt{(1-(u)^2)}}$$ where I can only think of substituting $u = sin(u)$ using the pythagorean theorem. Meaning that their(author) $du$ does not equate to $cosxdx$
$endgroup$
– oxodo
Dec 30 '18 at 0:37
$begingroup$
@oxodo In this case, I would actually make a new variable $z$ and say $u=sin z$. Then, you need to make the substitution using $du=cos zdz$. Try evaluating the integral in terms of $z$ and see what you get. Then, you can substitute $z=sin^{-1}u$ and $u=x-1$ back into your final solution.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:39
1
$begingroup$
Cheers. But why would you think of doing that? I mean since you have $int frac{du}{sqrt{(1-(u)^2)}}$ shouldn't the obvious thing to do be to recognize that $sqrt{(1-(u)^2})= cosu$? Thus leaving me with $frac{du}{cosu)}$ hence $int du(secu)$?
$endgroup$
– oxodo
Dec 30 '18 at 0:45
1
$begingroup$
However, $sqrt{1-u^2}neq cos u$. That equation is simply incorrect. I think you are basing this off $u=sin u$, which is also clearly incorrect. If $u=sin u$, then we could plug in $u=pi$ and get $pi=sin pi=0$, which is clearly a contradiction. This is why we have to make a new variable and say $u=sin z$ instead of $u=sin u$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:58
1
$begingroup$
@oxodo No, what is limiting the substitution $u=sin u$ is that it is a straight-out false equation. You can not just use the equation $u=sin u$ because it is simply not true. If you plug in any value of $u$ other than $u=0$ into that equation, you will get a contradiction ($u=pi$ was simply one example.) This is why you need to create a new variable and say $u=sin z$. Then, you can apply Pythagorean theorem to get $sqrt{1-u^2}=cos z$ and $du=cos zdz$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 1:17
|
show 1 more comment
1
$begingroup$
$$I=intfrac{dx}{sqrt{1-x^2}}$$
$$x=sin uRightarrow dx=cos u,du$$
$$Rightarrow I=int frac{cos u,du}{sqrt{1-sin^2u}}$$
Recalling that $$cos^2x=1-sin^2x$$
We have
$$I=intfrac{cos u,du}{cos u}$$
$$I=int du$$
$$I=u$$
$$I=arcsin x+C$$
answered Dec 30 '18 at 1:19
clathratusclathratus
5,2041439
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
$begingroup$
So, when you make the substitution $u=sin x$, you also need to make the substitution for $du$ as well:
$$du=frac{du}{dx}dx=cos xdx$$
Thus, the new integral becomes:
$$int frac{du}{sqrt{1-u^2}}=int frac{dxcos x}{sqrt{1-sin^2x}}$$
Hopefully, you can take it from there.
answered Dec 30 '18 at 0:26
Noble MushtakNoble Mushtak
15.4k1835
$endgroup$
$begingroup$
I do get that. I should've maybe added that the original integral was : $$int frac{du}{sqrt{(1-(x-1)^2)}}$$ . The author then substitued $(x-1)$ with $u$ giving me $du = dx$. We then get $$int frac{du}{sqrt{(1-(u)^2)}}$$ where I can only think of substituting $u = sin(u)$ using the pythagorean theorem. Meaning that their(author) $du$ does not equate to $cosxdx$
$endgroup$
– oxodo
Dec 30 '18 at 0:37
$begingroup$
@oxodo In this case, I would actually make a new variable $z$ and say $u=sin z$. Then, you need to make the substitution using $du=cos zdz$. Try evaluating the integral in terms of $z$ and see what you get. Then, you can substitute $z=sin^{-1}u$ and $u=x-1$ back into your final solution.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:39
1
$begingroup$
Cheers. But why would you think of doing that? I mean since you have $int frac{du}{sqrt{(1-(u)^2)}}$ shouldn't the obvious thing to do be to recognize that $sqrt{(1-(u)^2})= cosu$? Thus leaving me with $frac{du}{cosu)}$ hence $int du(secu)$?
$endgroup$
– oxodo
Dec 30 '18 at 0:45
1
$begingroup$
However, $sqrt{1-u^2}neq cos u$. That equation is simply incorrect. I think you are basing this off $u=sin u$, which is also clearly incorrect. If $u=sin u$, then we could plug in $u=pi$ and get $pi=sin pi=0$, which is clearly a contradiction. This is why we have to make a new variable and say $u=sin z$ instead of $u=sin u$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:58
1
$begingroup$
@oxodo No, what is limiting the substitution $u=sin u$ is that it is a straight-out false equation. You can not just use the equation $u=sin u$ because it is simply not true. If you plug in any value of $u$ other than $u=0$ into that equation, you will get a contradiction ($u=pi$ was simply one example.) This is why you need to create a new variable and say $u=sin z$. Then, you can apply Pythagorean theorem to get $sqrt{1-u^2}=cos z$ and $du=cos zdz$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 1:17
|
show 1 more comment
3
$begingroup$
So, when you make the substitution $u=sin x$, you also need to make the substitution for $du$ as well:
$$du=frac{du}{dx}dx=cos xdx$$
Thus, the new integral becomes:
$$int frac{du}{sqrt{1-u^2}}=int frac{dxcos x}{sqrt{1-sin^2x}}$$
Hopefully, you can take it from there.
answered Dec 30 '18 at 0:26
Noble MushtakNoble Mushtak
15.4k1835
$endgroup$
$begingroup$
I do get that. I should've maybe added that the original integral was : $$int frac{du}{sqrt{(1-(x-1)^2)}}$$ . The author then substitued $(x-1)$ with $u$ giving me $du = dx$. We then get $$int frac{du}{sqrt{(1-(u)^2)}}$$ where I can only think of substituting $u = sin(u)$ using the pythagorean theorem. Meaning that their(author) $du$ does not equate to $cosxdx$
$endgroup$
– oxodo
Dec 30 '18 at 0:37
$begingroup$
@oxodo In this case, I would actually make a new variable $z$ and say $u=sin z$. Then, you need to make the substitution using $du=cos zdz$. Try evaluating the integral in terms of $z$ and see what you get. Then, you can substitute $z=sin^{-1}u$ and $u=x-1$ back into your final solution.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:39
1
$begingroup$
Cheers. But why would you think of doing that? I mean since you have $int frac{du}{sqrt{(1-(u)^2)}}$ shouldn't the obvious thing to do be to recognize that $sqrt{(1-(u)^2})= cosu$? Thus leaving me with $frac{du}{cosu)}$ hence $int du(secu)$?
$endgroup$
– oxodo
Dec 30 '18 at 0:45
1
$begingroup$
However, $sqrt{1-u^2}neq cos u$. That equation is simply incorrect. I think you are basing this off $u=sin u$, which is also clearly incorrect. If $u=sin u$, then we could plug in $u=pi$ and get $pi=sin pi=0$, which is clearly a contradiction. This is why we have to make a new variable and say $u=sin z$ instead of $u=sin u$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:58
1
$begingroup$
@oxodo No, what is limiting the substitution $u=sin u$ is that it is a straight-out false equation. You can not just use the equation $u=sin u$ because it is simply not true. If you plug in any value of $u$ other than $u=0$ into that equation, you will get a contradiction ($u=pi$ was simply one example.) This is why you need to create a new variable and say $u=sin z$. Then, you can apply Pythagorean theorem to get $sqrt{1-u^2}=cos z$ and $du=cos zdz$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 1:17
|
show 1 more comment
3
3
3
$begingroup$
So, when you make the substitution $u=sin x$, you also need to make the substitution for $du$ as well:
$$du=frac{du}{dx}dx=cos xdx$$
Thus, the new integral becomes:
$$int frac{du}{sqrt{1-u^2}}=int frac{dxcos x}{sqrt{1-sin^2x}}$$
Hopefully, you can take it from there.
answered Dec 30 '18 at 0:26
Noble MushtakNoble Mushtak
15.4k1835
$endgroup$
So, when you make the substitution $u=sin x$, you also need to make the substitution for $du$ as well:
$$du=frac{du}{dx}dx=cos xdx$$
Thus, the new integral becomes:
$$int frac{du}{sqrt{1-u^2}}=int frac{dxcos x}{sqrt{1-sin^2x}}$$
Hopefully, you can take it from there.
answered Dec 30 '18 at 0:26
Noble MushtakNoble Mushtak
15.4k1835
answered Dec 30 '18 at 0:26
Noble MushtakNoble Mushtak
15.4k1835
answered Dec 30 '18 at 0:26
Noble MushtakNoble Mushtak
15.4k1835
answered Dec 30 '18 at 0:26
Noble MushtakNoble Mushtak
15.4k1835
15.4k1835
$begingroup$
I do get that. I should've maybe added that the original integral was : $$int frac{du}{sqrt{(1-(x-1)^2)}}$$ . The author then substitued $(x-1)$ with $u$ giving me $du = dx$. We then get $$int frac{du}{sqrt{(1-(u)^2)}}$$ where I can only think of substituting $u = sin(u)$ using the pythagorean theorem. Meaning that their(author) $du$ does not equate to $cosxdx$
$endgroup$
– oxodo
Dec 30 '18 at 0:37
$begingroup$
@oxodo In this case, I would actually make a new variable $z$ and say $u=sin z$. Then, you need to make the substitution using $du=cos zdz$. Try evaluating the integral in terms of $z$ and see what you get. Then, you can substitute $z=sin^{-1}u$ and $u=x-1$ back into your final solution.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:39
1
$begingroup$
Cheers. But why would you think of doing that? I mean since you have $int frac{du}{sqrt{(1-(u)^2)}}$ shouldn't the obvious thing to do be to recognize that $sqrt{(1-(u)^2})= cosu$? Thus leaving me with $frac{du}{cosu)}$ hence $int du(secu)$?
$endgroup$
– oxodo
Dec 30 '18 at 0:45
1
$begingroup$
However, $sqrt{1-u^2}neq cos u$. That equation is simply incorrect. I think you are basing this off $u=sin u$, which is also clearly incorrect. If $u=sin u$, then we could plug in $u=pi$ and get $pi=sin pi=0$, which is clearly a contradiction. This is why we have to make a new variable and say $u=sin z$ instead of $u=sin u$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:58
1
$begingroup$
@oxodo No, what is limiting the substitution $u=sin u$ is that it is a straight-out false equation. You can not just use the equation $u=sin u$ because it is simply not true. If you plug in any value of $u$ other than $u=0$ into that equation, you will get a contradiction ($u=pi$ was simply one example.) This is why you need to create a new variable and say $u=sin z$. Then, you can apply Pythagorean theorem to get $sqrt{1-u^2}=cos z$ and $du=cos zdz$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 1:17
|
show 1 more comment
$begingroup$
I do get that. I should've maybe added that the original integral was : $$int frac{du}{sqrt{(1-(x-1)^2)}}$$ . The author then substitued $(x-1)$ with $u$ giving me $du = dx$. We then get $$int frac{du}{sqrt{(1-(u)^2)}}$$ where I can only think of substituting $u = sin(u)$ using the pythagorean theorem. Meaning that their(author) $du$ does not equate to $cosxdx$
$endgroup$
– oxodo
Dec 30 '18 at 0:37
$begingroup$
@oxodo In this case, I would actually make a new variable $z$ and say $u=sin z$. Then, you need to make the substitution using $du=cos zdz$. Try evaluating the integral in terms of $z$ and see what you get. Then, you can substitute $z=sin^{-1}u$ and $u=x-1$ back into your final solution.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:39
1
$begingroup$
Cheers. But why would you think of doing that? I mean since you have $int frac{du}{sqrt{(1-(u)^2)}}$ shouldn't the obvious thing to do be to recognize that $sqrt{(1-(u)^2})= cosu$? Thus leaving me with $frac{du}{cosu)}$ hence $int du(secu)$?
$endgroup$
– oxodo
Dec 30 '18 at 0:45
1
$begingroup$
However, $sqrt{1-u^2}neq cos u$. That equation is simply incorrect. I think you are basing this off $u=sin u$, which is also clearly incorrect. If $u=sin u$, then we could plug in $u=pi$ and get $pi=sin pi=0$, which is clearly a contradiction. This is why we have to make a new variable and say $u=sin z$ instead of $u=sin u$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:58
1
$begingroup$
@oxodo No, what is limiting the substitution $u=sin u$ is that it is a straight-out false equation. You can not just use the equation $u=sin u$ because it is simply not true. If you plug in any value of $u$ other than $u=0$ into that equation, you will get a contradiction ($u=pi$ was simply one example.) This is why you need to create a new variable and say $u=sin z$. Then, you can apply Pythagorean theorem to get $sqrt{1-u^2}=cos z$ and $du=cos zdz$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 1:17
$begingroup$
I do get that. I should've maybe added that the original integral was : $$int frac{du}{sqrt{(1-(x-1)^2)}}$$ . The author then substitued $(x-1)$ with $u$ giving me $du = dx$. We then get $$int frac{du}{sqrt{(1-(u)^2)}}$$ where I can only think of substituting $u = sin(u)$ using the pythagorean theorem. Meaning that their(author) $du$ does not equate to $cosxdx$
$endgroup$
– oxodo
Dec 30 '18 at 0:37
$begingroup$
I do get that. I should've maybe added that the original integral was : $$int frac{du}{sqrt{(1-(x-1)^2)}}$$ . The author then substitued $(x-1)$ with $u$ giving me $du = dx$. We then get $$int frac{du}{sqrt{(1-(u)^2)}}$$ where I can only think of substituting $u = sin(u)$ using the pythagorean theorem. Meaning that their(author) $du$ does not equate to $cosxdx$
$endgroup$
– oxodo
Dec 30 '18 at 0:37
$begingroup$
@oxodo In this case, I would actually make a new variable $z$ and say $u=sin z$. Then, you need to make the substitution using $du=cos zdz$. Try evaluating the integral in terms of $z$ and see what you get. Then, you can substitute $z=sin^{-1}u$ and $u=x-1$ back into your final solution.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:39
$begingroup$
@oxodo In this case, I would actually make a new variable $z$ and say $u=sin z$. Then, you need to make the substitution using $du=cos zdz$. Try evaluating the integral in terms of $z$ and see what you get. Then, you can substitute $z=sin^{-1}u$ and $u=x-1$ back into your final solution.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:39
1
1
$begingroup$
Cheers. But why would you think of doing that? I mean since you have $int frac{du}{sqrt{(1-(u)^2)}}$ shouldn't the obvious thing to do be to recognize that $sqrt{(1-(u)^2})= cosu$? Thus leaving me with $frac{du}{cosu)}$ hence $int du(secu)$?
$endgroup$
– oxodo
Dec 30 '18 at 0:45
$begingroup$
Cheers. But why would you think of doing that? I mean since you have $int frac{du}{sqrt{(1-(u)^2)}}$ shouldn't the obvious thing to do be to recognize that $sqrt{(1-(u)^2})= cosu$? Thus leaving me with $frac{du}{cosu)}$ hence $int du(secu)$?
$endgroup$
– oxodo
Dec 30 '18 at 0:45
1
1
$begingroup$
However, $sqrt{1-u^2}neq cos u$. That equation is simply incorrect. I think you are basing this off $u=sin u$, which is also clearly incorrect. If $u=sin u$, then we could plug in $u=pi$ and get $pi=sin pi=0$, which is clearly a contradiction. This is why we have to make a new variable and say $u=sin z$ instead of $u=sin u$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:58
$begingroup$
However, $sqrt{1-u^2}neq cos u$. That equation is simply incorrect. I think you are basing this off $u=sin u$, which is also clearly incorrect. If $u=sin u$, then we could plug in $u=pi$ and get $pi=sin pi=0$, which is clearly a contradiction. This is why we have to make a new variable and say $u=sin z$ instead of $u=sin u$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 0:58
1
1
$begingroup$
@oxodo No, what is limiting the substitution $u=sin u$ is that it is a straight-out false equation. You can not just use the equation $u=sin u$ because it is simply not true. If you plug in any value of $u$ other than $u=0$ into that equation, you will get a contradiction ($u=pi$ was simply one example.) This is why you need to create a new variable and say $u=sin z$. Then, you can apply Pythagorean theorem to get $sqrt{1-u^2}=cos z$ and $du=cos zdz$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 1:17
$begingroup$
@oxodo No, what is limiting the substitution $u=sin u$ is that it is a straight-out false equation. You can not just use the equation $u=sin u$ because it is simply not true. If you plug in any value of $u$ other than $u=0$ into that equation, you will get a contradiction ($u=pi$ was simply one example.) This is why you need to create a new variable and say $u=sin z$. Then, you can apply Pythagorean theorem to get $sqrt{1-u^2}=cos z$ and $du=cos zdz$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 1:17
|
show 1 more comment
1
$begingroup$
$$I=intfrac{dx}{sqrt{1-x^2}}$$
$$x=sin uRightarrow dx=cos u,du$$
$$Rightarrow I=int frac{cos u,du}{sqrt{1-sin^2u}}$$
Recalling that $$cos^2x=1-sin^2x$$
We have
$$I=intfrac{cos u,du}{cos u}$$
$$I=int du$$
$$I=u$$
$$I=arcsin x+C$$
answered Dec 30 '18 at 1:19
clathratusclathratus
5,2041439
$endgroup$
add a comment |
1
$begingroup$
$$I=intfrac{dx}{sqrt{1-x^2}}$$
$$x=sin uRightarrow dx=cos u,du$$
$$Rightarrow I=int frac{cos u,du}{sqrt{1-sin^2u}}$$
Recalling that $$cos^2x=1-sin^2x$$
We have
$$I=intfrac{cos u,du}{cos u}$$
$$I=int du$$
$$I=u$$
$$I=arcsin x+C$$
answered Dec 30 '18 at 1:19
clathratusclathratus
5,2041439
$endgroup$
add a comment |
1
1
1
$begingroup$
$$I=intfrac{dx}{sqrt{1-x^2}}$$
$$x=sin uRightarrow dx=cos u,du$$
$$Rightarrow I=int frac{cos u,du}{sqrt{1-sin^2u}}$$
Recalling that $$cos^2x=1-sin^2x$$
We have
$$I=intfrac{cos u,du}{cos u}$$
$$I=int du$$
$$I=u$$
$$I=arcsin x+C$$
answered Dec 30 '18 at 1:19
clathratusclathratus
5,2041439
$endgroup$
$$I=intfrac{dx}{sqrt{1-x^2}}$$
$$x=sin uRightarrow dx=cos u,du$$
$$Rightarrow I=int frac{cos u,du}{sqrt{1-sin^2u}}$$
Recalling that $$cos^2x=1-sin^2x$$
We have
$$I=intfrac{cos u,du}{cos u}$$
$$I=int du$$
$$I=u$$
$$I=arcsin x+C$$
answered Dec 30 '18 at 1:19
clathratusclathratus
5,2041439
answered Dec 30 '18 at 1:19
clathratusclathratus
5,2041439
answered Dec 30 '18 at 1:19
clathratusclathratus
5,2041439
answered Dec 30 '18 at 1:19
clathratusclathratus
5,2041439
5,2041439
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