Mixing
$intfrac{sin(x)}{sqrt{1-cos^2(x)}}$ Where did I mess up the domain?
$begingroup$
So we have $$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx$$
My book uses a $u$-substitution with $u=cos(x)$, $du=-sin(x)$,and they get $$intfrac{-du}{sqrt{1-u^2}}$$ which gives them $arccos(u)+C=arccos(cos(x))+C$
I understand that the domain of $arccos(x)$ means that $arccos(cos(x))$ is only equal to $x$ on $[0,pi]$, and is equal to some shifted version of $x$ for other values. However, I solved the integral differently and it resulted in $x + C$, but that isn't correct. I know that somewhere I messed up the domain, but I can't tell where. Here is my solution.
$$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx$$
$1-cos^2(x)=sin^2(x)$ so $$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx=intfrac{sin(x)}{sqrt{sin^2(x)}}=intfrac{sin(x)}{sin(x)}dx=int{1cdot dx}$$
So I get $x + C$ as my answer. Where did I mess up?
indefinite-integrals trigonometric-integrals
edited Jan 29 at 13:00
Thomas Shelby
4,5862727
asked Jan 29 at 12:49
Ibrahim ElsadekIbrahim Elsadek
112
$endgroup$
add a comment |
$begingroup$
So we have $$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx$$
My book uses a $u$-substitution with $u=cos(x)$, $du=-sin(x)$,and they get $$intfrac{-du}{sqrt{1-u^2}}$$ which gives them $arccos(u)+C=arccos(cos(x))+C$
I understand that the domain of $arccos(x)$ means that $arccos(cos(x))$ is only equal to $x$ on $[0,pi]$, and is equal to some shifted version of $x$ for other values. However, I solved the integral differently and it resulted in $x + C$, but that isn't correct. I know that somewhere I messed up the domain, but I can't tell where. Here is my solution.
$$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx$$
$1-cos^2(x)=sin^2(x)$ so $$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx=intfrac{sin(x)}{sqrt{sin^2(x)}}=intfrac{sin(x)}{sin(x)}dx=int{1cdot dx}$$
So I get $x + C$ as my answer. Where did I mess up?
indefinite-integrals trigonometric-integrals
edited Jan 29 at 13:00
Thomas Shelby
4,5862727
asked Jan 29 at 12:49
Ibrahim ElsadekIbrahim Elsadek
112
$endgroup$
$begingroup$
What makes you think that it is wrong ? And why do you incriminate the domain ?
$endgroup$
– Yves Daoust
Jan 29 at 12:56
add a comment |
$begingroup$
So we have $$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx$$
My book uses a $u$-substitution with $u=cos(x)$, $du=-sin(x)$,and they get $$intfrac{-du}{sqrt{1-u^2}}$$ which gives them $arccos(u)+C=arccos(cos(x))+C$
I understand that the domain of $arccos(x)$ means that $arccos(cos(x))$ is only equal to $x$ on $[0,pi]$, and is equal to some shifted version of $x$ for other values. However, I solved the integral differently and it resulted in $x + C$, but that isn't correct. I know that somewhere I messed up the domain, but I can't tell where. Here is my solution.
$$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx$$
$1-cos^2(x)=sin^2(x)$ so $$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx=intfrac{sin(x)}{sqrt{sin^2(x)}}=intfrac{sin(x)}{sin(x)}dx=int{1cdot dx}$$
So I get $x + C$ as my answer. Where did I mess up?
indefinite-integrals trigonometric-integrals
edited Jan 29 at 13:00
Thomas Shelby
4,5862727
asked Jan 29 at 12:49
Ibrahim ElsadekIbrahim Elsadek
112
$endgroup$
So we have $$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx$$
My book uses a $u$-substitution with $u=cos(x)$, $du=-sin(x)$,and they get $$intfrac{-du}{sqrt{1-u^2}}$$ which gives them $arccos(u)+C=arccos(cos(x))+C$
I understand that the domain of $arccos(x)$ means that $arccos(cos(x))$ is only equal to $x$ on $[0,pi]$, and is equal to some shifted version of $x$ for other values. However, I solved the integral differently and it resulted in $x + C$, but that isn't correct. I know that somewhere I messed up the domain, but I can't tell where. Here is my solution.
$$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx$$
$1-cos^2(x)=sin^2(x)$ so $$intfrac{sin(x)}{sqrt{1-cos^2(x)}}dx=intfrac{sin(x)}{sqrt{sin^2(x)}}=intfrac{sin(x)}{sin(x)}dx=int{1cdot dx}$$
So I get $x + C$ as my answer. Where did I mess up?
indefinite-integrals trigonometric-integrals
indefinite-integrals trigonometric-integrals
edited Jan 29 at 13:00
Thomas Shelby
4,5862727
asked Jan 29 at 12:49
Ibrahim ElsadekIbrahim Elsadek
112
edited Jan 29 at 13:00
Thomas Shelby
4,5862727
asked Jan 29 at 12:49
Ibrahim ElsadekIbrahim Elsadek
112
edited Jan 29 at 13:00
Thomas Shelby
4,5862727
edited Jan 29 at 13:00
Thomas Shelby
4,5862727
edited Jan 29 at 13:00
Thomas Shelby
4,5862727
4,5862727
asked Jan 29 at 12:49
Ibrahim ElsadekIbrahim Elsadek
112
asked Jan 29 at 12:49
Ibrahim ElsadekIbrahim Elsadek
112
asked Jan 29 at 12:49
Ibrahim ElsadekIbrahim Elsadek
112
112
$begingroup$
What makes you think that it is wrong ? And why do you incriminate the domain ?
$endgroup$
– Yves Daoust
Jan 29 at 12:56
add a comment |
$begingroup$
What makes you think that it is wrong ? And why do you incriminate the domain ?
$endgroup$
– Yves Daoust
Jan 29 at 12:56
$begingroup$
What makes you think that it is wrong ? And why do you incriminate the domain ?
$endgroup$
– Yves Daoust
Jan 29 at 12:56
$begingroup$
What makes you think that it is wrong ? And why do you incriminate the domain ?
$endgroup$
– Yves Daoust
Jan 29 at 12:56
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You messed up because $sqrt{sin^2(x)} =sin x$ is not true for all values of $x$. In fact, it is only true if $sin(x)geq0$. If $sin(x)<0$, then $sqrt{sin^2(x)}=-sin(x)$. In general, $sqrt{sin^2(x)} = |sin x|$.
answered Jan 29 at 12:53
5xum5xum
91.9k394161
$endgroup$
$begingroup$
OK, so if i were to do $intfrac{sinx}{|{sinx}|}dx$ that would be right? I guess I have to do u-sub because I don't know how to integrate absolute values. Thanks
$endgroup$
– Ibrahim Elsadek
Jan 29 at 13:02
add a comment |
$begingroup$
We have $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=1 iff sin(x) >0$
and $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=-1 iff sin(x) <0$.
Can you take it from here ?
answered Jan 29 at 12:54
FredFred
48.9k11849
$endgroup$
add a comment |
$begingroup$
The integrand is a square wave. Its antiderivative is a triangle wave.
answered Jan 29 at 13:03
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3092128%2fint-frac-sinx-sqrt1-cos2x-where-did-i-mess-up-the-domain%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You messed up because $sqrt{sin^2(x)} =sin x$ is not true for all values of $x$. In fact, it is only true if $sin(x)geq0$. If $sin(x)<0$, then $sqrt{sin^2(x)}=-sin(x)$. In general, $sqrt{sin^2(x)} = |sin x|$.
answered Jan 29 at 12:53
5xum5xum
91.9k394161
$endgroup$
$begingroup$
OK, so if i were to do $intfrac{sinx}{|{sinx}|}dx$ that would be right? I guess I have to do u-sub because I don't know how to integrate absolute values. Thanks
$endgroup$
– Ibrahim Elsadek
Jan 29 at 13:02
add a comment |
$begingroup$
You messed up because $sqrt{sin^2(x)} =sin x$ is not true for all values of $x$. In fact, it is only true if $sin(x)geq0$. If $sin(x)<0$, then $sqrt{sin^2(x)}=-sin(x)$. In general, $sqrt{sin^2(x)} = |sin x|$.
answered Jan 29 at 12:53
5xum5xum
91.9k394161
$endgroup$
$begingroup$
OK, so if i were to do $intfrac{sinx}{|{sinx}|}dx$ that would be right? I guess I have to do u-sub because I don't know how to integrate absolute values. Thanks
$endgroup$
– Ibrahim Elsadek
Jan 29 at 13:02
add a comment |
$begingroup$
You messed up because $sqrt{sin^2(x)} =sin x$ is not true for all values of $x$. In fact, it is only true if $sin(x)geq0$. If $sin(x)<0$, then $sqrt{sin^2(x)}=-sin(x)$. In general, $sqrt{sin^2(x)} = |sin x|$.
answered Jan 29 at 12:53
5xum5xum
91.9k394161
$endgroup$
You messed up because $sqrt{sin^2(x)} =sin x$ is not true for all values of $x$. In fact, it is only true if $sin(x)geq0$. If $sin(x)<0$, then $sqrt{sin^2(x)}=-sin(x)$. In general, $sqrt{sin^2(x)} = |sin x|$.
answered Jan 29 at 12:53
5xum5xum
91.9k394161
answered Jan 29 at 12:53
5xum5xum
91.9k394161
answered Jan 29 at 12:53
5xum5xum
91.9k394161
answered Jan 29 at 12:53
5xum5xum
91.9k394161
91.9k394161
$begingroup$
OK, so if i were to do $intfrac{sinx}{|{sinx}|}dx$ that would be right? I guess I have to do u-sub because I don't know how to integrate absolute values. Thanks
$endgroup$
– Ibrahim Elsadek
Jan 29 at 13:02
add a comment |
$begingroup$
OK, so if i were to do $intfrac{sinx}{|{sinx}|}dx$ that would be right? I guess I have to do u-sub because I don't know how to integrate absolute values. Thanks
$endgroup$
– Ibrahim Elsadek
Jan 29 at 13:02
$begingroup$
OK, so if i were to do $intfrac{sinx}{|{sinx}|}dx$ that would be right? I guess I have to do u-sub because I don't know how to integrate absolute values. Thanks
$endgroup$
– Ibrahim Elsadek
Jan 29 at 13:02
$begingroup$
OK, so if i were to do $intfrac{sinx}{|{sinx}|}dx$ that would be right? I guess I have to do u-sub because I don't know how to integrate absolute values. Thanks
$endgroup$
– Ibrahim Elsadek
Jan 29 at 13:02
add a comment |
$begingroup$
We have $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=1 iff sin(x) >0$
and $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=-1 iff sin(x) <0$.
Can you take it from here ?
answered Jan 29 at 12:54
FredFred
48.9k11849
$endgroup$
add a comment |
$begingroup$
We have $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=1 iff sin(x) >0$
and $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=-1 iff sin(x) <0$.
Can you take it from here ?
answered Jan 29 at 12:54
FredFred
48.9k11849
$endgroup$
add a comment |
$begingroup$
We have $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=1 iff sin(x) >0$
and $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=-1 iff sin(x) <0$.
Can you take it from here ?
answered Jan 29 at 12:54
FredFred
48.9k11849
$endgroup$
We have $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=1 iff sin(x) >0$
and $frac{sin(x)}{sqrt{sin^2(x)}}=frac{sin(x)}{|sin(x)|}=-1 iff sin(x) <0$.
Can you take it from here ?
answered Jan 29 at 12:54
FredFred
48.9k11849
answered Jan 29 at 12:54
FredFred
48.9k11849
answered Jan 29 at 12:54
FredFred
48.9k11849
answered Jan 29 at 12:54
FredFred
48.9k11849
48.9k11849
add a comment |
add a comment |
$begingroup$
The integrand is a square wave. Its antiderivative is a triangle wave.
answered Jan 29 at 13:03
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
The integrand is a square wave. Its antiderivative is a triangle wave.
answered Jan 29 at 13:03
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
The integrand is a square wave. Its antiderivative is a triangle wave.
answered Jan 29 at 13:03
Yves DaoustYves Daoust
131k676229
$endgroup$
The integrand is a square wave. Its antiderivative is a triangle wave.
answered Jan 29 at 13:03
Yves DaoustYves Daoust
131k676229
answered Jan 29 at 13:03
Yves DaoustYves Daoust
131k676229
answered Jan 29 at 13:03
Yves DaoustYves Daoust
131k676229
answered Jan 29 at 13:03
Yves DaoustYves Daoust
131k676229
131k676229
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3092128%2fint-frac-sinx-sqrt1-cos2x-where-did-i-mess-up-the-domain%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
What makes you think that it is wrong ? And why do you incriminate the domain ?
$endgroup$
– Yves Daoust
Jan 29 at 12:56