Mixing
What substitution should I notice in this integral?
$begingroup$
I've solved this integral by partial integration method ('u' is first fraction and dv is the second fraction), but I've been told that there is a much simpler method using substitution which I can't see.I've tried to substitute arcsin(x) which eliminates the root and leaves me with sin functions in the integral (it didn't help).
$$int frac {arcsin(x)}{sqrt {1-x^2}} * frac {1+x^2}{x^2} dx$$
indefinite-integrals
asked Jan 30 at 17:19
JoeDoughJoeDough
226
$endgroup$
add a comment |
$begingroup$
I've solved this integral by partial integration method ('u' is first fraction and dv is the second fraction), but I've been told that there is a much simpler method using substitution which I can't see.I've tried to substitute arcsin(x) which eliminates the root and leaves me with sin functions in the integral (it didn't help).
$$int frac {arcsin(x)}{sqrt {1-x^2}} * frac {1+x^2}{x^2} dx$$
indefinite-integrals
asked Jan 30 at 17:19
JoeDoughJoeDough
226
$endgroup$
add a comment |
$begingroup$
I've solved this integral by partial integration method ('u' is first fraction and dv is the second fraction), but I've been told that there is a much simpler method using substitution which I can't see.I've tried to substitute arcsin(x) which eliminates the root and leaves me with sin functions in the integral (it didn't help).
$$int frac {arcsin(x)}{sqrt {1-x^2}} * frac {1+x^2}{x^2} dx$$
indefinite-integrals
asked Jan 30 at 17:19
JoeDoughJoeDough
226
$endgroup$
I've solved this integral by partial integration method ('u' is first fraction and dv is the second fraction), but I've been told that there is a much simpler method using substitution which I can't see.I've tried to substitute arcsin(x) which eliminates the root and leaves me with sin functions in the integral (it didn't help).
$$int frac {arcsin(x)}{sqrt {1-x^2}} * frac {1+x^2}{x^2} dx$$
indefinite-integrals
indefinite-integrals
asked Jan 30 at 17:19
JoeDoughJoeDough
226
asked Jan 30 at 17:19
JoeDoughJoeDough
226
asked Jan 30 at 17:19
JoeDoughJoeDough
226
asked Jan 30 at 17:19
JoeDoughJoeDough
226
asked Jan 30 at 17:19
JoeDoughJoeDough
226
226
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You're going to have to use integration by parts at some point. But as Harnak said, we can use $u$ substituion to have a much easier time. Let $u = arcsin(x)$. Then
begin{align*}
int frac{arcsin(x)}{sqrt{1-x^2}} (frac{1+x^2}{x^2}) dx &= int u(1+csc^2(u))du \
&=frac{u^2}{2} +int ucsc^2(u) du \
&= frac{u^2}{2} - ucot(u) + int cot(u)du \
&= frac{u^2}{2} - ucot(u)+lnmidsin(u)mid+C
end{align*}
From this point, make the necessary substitutions back in and simplify.
answered Jan 30 at 18:06
HyperionHyperion
702111
$endgroup$
add a comment |
$begingroup$
Hint: try the following substitution $$t= arcsin(x)$$
and remember the derivative of arcsine.
answered Jan 30 at 17:26
HarnakHarnak
1,369512
$endgroup$
$begingroup$
Are you sure it works ?
$endgroup$
– hamam_Abdallah
Jan 30 at 17:30
$begingroup$
It works according to wolframalpha.I've already tried this method, this gives me integral t* (1+sin^2 (t))/ sin^2 (t).I'm not sure how to proceed from here.Maybe trigonometric substitutions (t=tan(x/2))?
$endgroup$
– JoeDough
Jan 30 at 17:56
add a comment |
Your Answer
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%2f3093822%2fwhat-substitution-should-i-notice-in-this-integral%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You're going to have to use integration by parts at some point. But as Harnak said, we can use $u$ substituion to have a much easier time. Let $u = arcsin(x)$. Then
begin{align*}
int frac{arcsin(x)}{sqrt{1-x^2}} (frac{1+x^2}{x^2}) dx &= int u(1+csc^2(u))du \
&=frac{u^2}{2} +int ucsc^2(u) du \
&= frac{u^2}{2} - ucot(u) + int cot(u)du \
&= frac{u^2}{2} - ucot(u)+lnmidsin(u)mid+C
end{align*}
From this point, make the necessary substitutions back in and simplify.
answered Jan 30 at 18:06
HyperionHyperion
702111
$endgroup$
add a comment |
$begingroup$
You're going to have to use integration by parts at some point. But as Harnak said, we can use $u$ substituion to have a much easier time. Let $u = arcsin(x)$. Then
begin{align*}
int frac{arcsin(x)}{sqrt{1-x^2}} (frac{1+x^2}{x^2}) dx &= int u(1+csc^2(u))du \
&=frac{u^2}{2} +int ucsc^2(u) du \
&= frac{u^2}{2} - ucot(u) + int cot(u)du \
&= frac{u^2}{2} - ucot(u)+lnmidsin(u)mid+C
end{align*}
From this point, make the necessary substitutions back in and simplify.
answered Jan 30 at 18:06
HyperionHyperion
702111
$endgroup$
add a comment |
$begingroup$
You're going to have to use integration by parts at some point. But as Harnak said, we can use $u$ substituion to have a much easier time. Let $u = arcsin(x)$. Then
begin{align*}
int frac{arcsin(x)}{sqrt{1-x^2}} (frac{1+x^2}{x^2}) dx &= int u(1+csc^2(u))du \
&=frac{u^2}{2} +int ucsc^2(u) du \
&= frac{u^2}{2} - ucot(u) + int cot(u)du \
&= frac{u^2}{2} - ucot(u)+lnmidsin(u)mid+C
end{align*}
From this point, make the necessary substitutions back in and simplify.
answered Jan 30 at 18:06
HyperionHyperion
702111
$endgroup$
You're going to have to use integration by parts at some point. But as Harnak said, we can use $u$ substituion to have a much easier time. Let $u = arcsin(x)$. Then
begin{align*}
int frac{arcsin(x)}{sqrt{1-x^2}} (frac{1+x^2}{x^2}) dx &= int u(1+csc^2(u))du \
&=frac{u^2}{2} +int ucsc^2(u) du \
&= frac{u^2}{2} - ucot(u) + int cot(u)du \
&= frac{u^2}{2} - ucot(u)+lnmidsin(u)mid+C
end{align*}
From this point, make the necessary substitutions back in and simplify.
answered Jan 30 at 18:06
HyperionHyperion
702111
answered Jan 30 at 18:06
HyperionHyperion
702111
answered Jan 30 at 18:06
HyperionHyperion
702111
answered Jan 30 at 18:06
HyperionHyperion
702111
702111
add a comment |
add a comment |
$begingroup$
Hint: try the following substitution $$t= arcsin(x)$$
and remember the derivative of arcsine.
answered Jan 30 at 17:26
HarnakHarnak
1,369512
$endgroup$
$begingroup$
Are you sure it works ?
$endgroup$
– hamam_Abdallah
Jan 30 at 17:30
$begingroup$
It works according to wolframalpha.I've already tried this method, this gives me integral t* (1+sin^2 (t))/ sin^2 (t).I'm not sure how to proceed from here.Maybe trigonometric substitutions (t=tan(x/2))?
$endgroup$
– JoeDough
Jan 30 at 17:56
add a comment |
$begingroup$
Hint: try the following substitution $$t= arcsin(x)$$
and remember the derivative of arcsine.
answered Jan 30 at 17:26
HarnakHarnak
1,369512
$endgroup$
$begingroup$
Are you sure it works ?
$endgroup$
– hamam_Abdallah
Jan 30 at 17:30
$begingroup$
It works according to wolframalpha.I've already tried this method, this gives me integral t* (1+sin^2 (t))/ sin^2 (t).I'm not sure how to proceed from here.Maybe trigonometric substitutions (t=tan(x/2))?
$endgroup$
– JoeDough
Jan 30 at 17:56
add a comment |
$begingroup$
Hint: try the following substitution $$t= arcsin(x)$$
and remember the derivative of arcsine.
answered Jan 30 at 17:26
HarnakHarnak
1,369512
$endgroup$
Hint: try the following substitution $$t= arcsin(x)$$
and remember the derivative of arcsine.
answered Jan 30 at 17:26
HarnakHarnak
1,369512
answered Jan 30 at 17:26
HarnakHarnak
1,369512
answered Jan 30 at 17:26
HarnakHarnak
1,369512
answered Jan 30 at 17:26
HarnakHarnak
1,369512
1,369512
$begingroup$
Are you sure it works ?
$endgroup$
– hamam_Abdallah
Jan 30 at 17:30
$begingroup$
It works according to wolframalpha.I've already tried this method, this gives me integral t* (1+sin^2 (t))/ sin^2 (t).I'm not sure how to proceed from here.Maybe trigonometric substitutions (t=tan(x/2))?
$endgroup$
– JoeDough
Jan 30 at 17:56
add a comment |
$begingroup$
Are you sure it works ?
$endgroup$
– hamam_Abdallah
Jan 30 at 17:30
$begingroup$
It works according to wolframalpha.I've already tried this method, this gives me integral t* (1+sin^2 (t))/ sin^2 (t).I'm not sure how to proceed from here.Maybe trigonometric substitutions (t=tan(x/2))?
$endgroup$
– JoeDough
Jan 30 at 17:56
$begingroup$
Are you sure it works ?
$endgroup$
– hamam_Abdallah
Jan 30 at 17:30
$begingroup$
Are you sure it works ?
$endgroup$
– hamam_Abdallah
Jan 30 at 17:30
$begingroup$
It works according to wolframalpha.I've already tried this method, this gives me integral t* (1+sin^2 (t))/ sin^2 (t).I'm not sure how to proceed from here.Maybe trigonometric substitutions (t=tan(x/2))?
$endgroup$
– JoeDough
Jan 30 at 17:56
$begingroup$
It works according to wolframalpha.I've already tried this method, this gives me integral t* (1+sin^2 (t))/ sin^2 (t).I'm not sure how to proceed from here.Maybe trigonometric substitutions (t=tan(x/2))?
$endgroup$
– JoeDough
Jan 30 at 17:56
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%2f3093822%2fwhat-substitution-should-i-notice-in-this-integral%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
