$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)},dx$
$begingroup$
I have been trying to solve the integral below using contour integration.
$$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)}dx, quad a>1.$$
I'd appreciate to see how you would solve it, since the fact that it goes to only $pi$ and not to $2pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.
complex-analysis contour-integration
$endgroup$
add a comment |
$begingroup$
I have been trying to solve the integral below using contour integration.
$$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)}dx, quad a>1.$$
I'd appreciate to see how you would solve it, since the fact that it goes to only $pi$ and not to $2pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.
complex-analysis contour-integration
$endgroup$
1
$begingroup$
I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
$endgroup$
– Robert Israel
Jan 3 at 16:33
$begingroup$
Oh true, ill give that a try. Thanks
$endgroup$
– JustANoob
Jan 3 at 16:39
$begingroup$
Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 16:42
add a comment |
$begingroup$
I have been trying to solve the integral below using contour integration.
$$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)}dx, quad a>1.$$
I'd appreciate to see how you would solve it, since the fact that it goes to only $pi$ and not to $2pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.
complex-analysis contour-integration
$endgroup$
I have been trying to solve the integral below using contour integration.
$$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)}dx, quad a>1.$$
I'd appreciate to see how you would solve it, since the fact that it goes to only $pi$ and not to $2pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.
complex-analysis contour-integration
complex-analysis contour-integration
edited Jan 3 at 18:36
Bernard
122k741116
122k741116
asked Jan 3 at 16:32
JustANoobJustANoob
399114
399114
1
$begingroup$
I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
$endgroup$
– Robert Israel
Jan 3 at 16:33
$begingroup$
Oh true, ill give that a try. Thanks
$endgroup$
– JustANoob
Jan 3 at 16:39
$begingroup$
Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 16:42
add a comment |
1
$begingroup$
I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
$endgroup$
– Robert Israel
Jan 3 at 16:33
$begingroup$
Oh true, ill give that a try. Thanks
$endgroup$
– JustANoob
Jan 3 at 16:39
$begingroup$
Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 16:42
1
1
$begingroup$
I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
$endgroup$
– Robert Israel
Jan 3 at 16:33
$begingroup$
I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
$endgroup$
– Robert Israel
Jan 3 at 16:33
$begingroup$
Oh true, ill give that a try. Thanks
$endgroup$
– JustANoob
Jan 3 at 16:39
$begingroup$
Oh true, ill give that a try. Thanks
$endgroup$
– JustANoob
Jan 3 at 16:39
$begingroup$
Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 16:42
$begingroup$
Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 16:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The chances of me saying this were high: let us avoid contour integration.
By symmetry
$$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
equals
$$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
or
$$ frac{pi}{a+sqrt{a^2-1}} $$
by partial fraction decomposition.
$endgroup$
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%2f3060749%2fint-0-pi-frac-sin2xa-cosx-dx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The chances of me saying this were high: let us avoid contour integration.
By symmetry
$$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
equals
$$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
or
$$ frac{pi}{a+sqrt{a^2-1}} $$
by partial fraction decomposition.
$endgroup$
add a comment |
$begingroup$
The chances of me saying this were high: let us avoid contour integration.
By symmetry
$$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
equals
$$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
or
$$ frac{pi}{a+sqrt{a^2-1}} $$
by partial fraction decomposition.
$endgroup$
add a comment |
$begingroup$
The chances of me saying this were high: let us avoid contour integration.
By symmetry
$$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
equals
$$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
or
$$ frac{pi}{a+sqrt{a^2-1}} $$
by partial fraction decomposition.
$endgroup$
The chances of me saying this were high: let us avoid contour integration.
By symmetry
$$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
equals
$$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
or
$$ frac{pi}{a+sqrt{a^2-1}} $$
by partial fraction decomposition.
answered Jan 3 at 16:51
Jack D'AurizioJack D'Aurizio
291k33284667
291k33284667
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%2f3060749%2fint-0-pi-frac-sin2xa-cosx-dx%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
1
$begingroup$
I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
$endgroup$
– Robert Israel
Jan 3 at 16:33
$begingroup$
Oh true, ill give that a try. Thanks
$endgroup$
– JustANoob
Jan 3 at 16:39
$begingroup$
Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 16:42