Integration of $int_0^frac{pi}{6} cos^{-3}2x sin2x , dx $?
0
$begingroup$
I tried substituting $x=frac{cos t}{2}$
but I didn't got anywhere.
Thanks!
calculus integration definite-integrals
asked Dec 11 '18 at 16:38
RaghavRaghav
557
$endgroup$
add a comment |
0
$begingroup$
I tried substituting $x=frac{cos t}{2}$
but I didn't got anywhere.
Thanks!
calculus integration definite-integrals
asked Dec 11 '18 at 16:38
RaghavRaghav
557
$endgroup$
add a comment |
0
0
0
$begingroup$
I tried substituting $x=frac{cos t}{2}$
but I didn't got anywhere.
Thanks!
calculus integration definite-integrals
asked Dec 11 '18 at 16:38
RaghavRaghav
557
$endgroup$
I tried substituting $x=frac{cos t}{2}$
but I didn't got anywhere.
Thanks!
calculus integration definite-integrals
calculus integration definite-integrals
asked Dec 11 '18 at 16:38
RaghavRaghav
557
asked Dec 11 '18 at 16:38
RaghavRaghav
557
asked Dec 11 '18 at 16:38
RaghavRaghav
557
asked Dec 11 '18 at 16:38
RaghavRaghav
557
asked Dec 11 '18 at 16:38
RaghavRaghav
557
557
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
4
$begingroup$
HINT: Rewrite the integrand as
$$cos^{-3}(2x)sin(2x)=tan(2x)sec^2(2x)$$
answered Dec 11 '18 at 16:40
FrpzzdFrpzzd
23k841109
$endgroup$
$begingroup$
wait, is it cos inverse cube x or $frac{1}{sec^3 x}$?
$endgroup$
– Raghav
Dec 11 '18 at 16:43
$begingroup$
Do you mean the integrand is $arccos^3 2x ,sin 2x$?
$endgroup$
– Bernard
Dec 11 '18 at 16:46
$begingroup$
It should be $sec^3x$, otherwise it would have said $(cos^{-1}x)^3$
$endgroup$
– Shubham Johri
Dec 11 '18 at 16:47
$begingroup$
@Bernard Yeah I meant that.
$endgroup$
– Raghav
Dec 11 '18 at 16:47
$begingroup$
@ShubhamJohri Ohh, ok. It's the first time I've encountered it, so I got confused.Thanks.
$endgroup$
– Raghav
Dec 11 '18 at 16:48
|
show 2 more comments
0
$begingroup$
Hint:
For $$intcos^m(2x)sin^{2n+1}(2x) dx,$$
choose $cos(2x)=y$
answered Dec 11 '18 at 17:33
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
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
});
}
});
draft saved
draft discarded
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%2f3035478%2fintegration-of-int-0-frac-pi6-cos-32x-sin2x-dx%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
4
$begingroup$
HINT: Rewrite the integrand as
$$cos^{-3}(2x)sin(2x)=tan(2x)sec^2(2x)$$
answered Dec 11 '18 at 16:40
FrpzzdFrpzzd
23k841109
$endgroup$
$begingroup$
wait, is it cos inverse cube x or $frac{1}{sec^3 x}$?
$endgroup$
– Raghav
Dec 11 '18 at 16:43
$begingroup$
Do you mean the integrand is $arccos^3 2x ,sin 2x$?
$endgroup$
– Bernard
Dec 11 '18 at 16:46
$begingroup$
It should be $sec^3x$, otherwise it would have said $(cos^{-1}x)^3$
$endgroup$
– Shubham Johri
Dec 11 '18 at 16:47
$begingroup$
@Bernard Yeah I meant that.
$endgroup$
– Raghav
Dec 11 '18 at 16:47
$begingroup$
@ShubhamJohri Ohh, ok. It's the first time I've encountered it, so I got confused.Thanks.
$endgroup$
– Raghav
Dec 11 '18 at 16:48
|
show 2 more comments
4
$begingroup$
HINT: Rewrite the integrand as
$$cos^{-3}(2x)sin(2x)=tan(2x)sec^2(2x)$$
answered Dec 11 '18 at 16:40
FrpzzdFrpzzd
23k841109
$endgroup$
$begingroup$
wait, is it cos inverse cube x or $frac{1}{sec^3 x}$?
$endgroup$
– Raghav
Dec 11 '18 at 16:43
$begingroup$
Do you mean the integrand is $arccos^3 2x ,sin 2x$?
$endgroup$
– Bernard
Dec 11 '18 at 16:46
$begingroup$
It should be $sec^3x$, otherwise it would have said $(cos^{-1}x)^3$
$endgroup$
– Shubham Johri
Dec 11 '18 at 16:47
$begingroup$
@Bernard Yeah I meant that.
$endgroup$
– Raghav
Dec 11 '18 at 16:47
$begingroup$
@ShubhamJohri Ohh, ok. It's the first time I've encountered it, so I got confused.Thanks.
$endgroup$
– Raghav
Dec 11 '18 at 16:48
|
show 2 more comments
4
4
4
$begingroup$
HINT: Rewrite the integrand as
$$cos^{-3}(2x)sin(2x)=tan(2x)sec^2(2x)$$
answered Dec 11 '18 at 16:40
FrpzzdFrpzzd
23k841109
$endgroup$
HINT: Rewrite the integrand as
$$cos^{-3}(2x)sin(2x)=tan(2x)sec^2(2x)$$
answered Dec 11 '18 at 16:40
FrpzzdFrpzzd
23k841109
answered Dec 11 '18 at 16:40
FrpzzdFrpzzd
23k841109
answered Dec 11 '18 at 16:40
FrpzzdFrpzzd
23k841109
answered Dec 11 '18 at 16:40
FrpzzdFrpzzd
23k841109
23k841109
$begingroup$
wait, is it cos inverse cube x or $frac{1}{sec^3 x}$?
$endgroup$
– Raghav
Dec 11 '18 at 16:43
$begingroup$
Do you mean the integrand is $arccos^3 2x ,sin 2x$?
$endgroup$
– Bernard
Dec 11 '18 at 16:46
$begingroup$
It should be $sec^3x$, otherwise it would have said $(cos^{-1}x)^3$
$endgroup$
– Shubham Johri
Dec 11 '18 at 16:47
$begingroup$
@Bernard Yeah I meant that.
$endgroup$
– Raghav
Dec 11 '18 at 16:47
$begingroup$
@ShubhamJohri Ohh, ok. It's the first time I've encountered it, so I got confused.Thanks.
$endgroup$
– Raghav
Dec 11 '18 at 16:48
|
show 2 more comments
$begingroup$
wait, is it cos inverse cube x or $frac{1}{sec^3 x}$?
$endgroup$
– Raghav
Dec 11 '18 at 16:43
$begingroup$
Do you mean the integrand is $arccos^3 2x ,sin 2x$?
$endgroup$
– Bernard
Dec 11 '18 at 16:46
$begingroup$
It should be $sec^3x$, otherwise it would have said $(cos^{-1}x)^3$
$endgroup$
– Shubham Johri
Dec 11 '18 at 16:47
$begingroup$
@Bernard Yeah I meant that.
$endgroup$
– Raghav
Dec 11 '18 at 16:47
$begingroup$
@ShubhamJohri Ohh, ok. It's the first time I've encountered it, so I got confused.Thanks.
$endgroup$
– Raghav
Dec 11 '18 at 16:48
$begingroup$
wait, is it cos inverse cube x or $frac{1}{sec^3 x}$?
$endgroup$
– Raghav
Dec 11 '18 at 16:43
$begingroup$
wait, is it cos inverse cube x or $frac{1}{sec^3 x}$?
$endgroup$
– Raghav
Dec 11 '18 at 16:43
$begingroup$
Do you mean the integrand is $arccos^3 2x ,sin 2x$?
$endgroup$
– Bernard
Dec 11 '18 at 16:46
$begingroup$
Do you mean the integrand is $arccos^3 2x ,sin 2x$?
$endgroup$
– Bernard
Dec 11 '18 at 16:46
$begingroup$
It should be $sec^3x$, otherwise it would have said $(cos^{-1}x)^3$
$endgroup$
– Shubham Johri
Dec 11 '18 at 16:47
$begingroup$
It should be $sec^3x$, otherwise it would have said $(cos^{-1}x)^3$
$endgroup$
– Shubham Johri
Dec 11 '18 at 16:47
$begingroup$
@Bernard Yeah I meant that.
$endgroup$
– Raghav
Dec 11 '18 at 16:47
$begingroup$
@Bernard Yeah I meant that.
$endgroup$
– Raghav
Dec 11 '18 at 16:47
$begingroup$
@ShubhamJohri Ohh, ok. It's the first time I've encountered it, so I got confused.Thanks.
$endgroup$
– Raghav
Dec 11 '18 at 16:48
$begingroup$
@ShubhamJohri Ohh, ok. It's the first time I've encountered it, so I got confused.Thanks.
$endgroup$
– Raghav
Dec 11 '18 at 16:48
|
show 2 more comments
0
$begingroup$
Hint:
For $$intcos^m(2x)sin^{2n+1}(2x) dx,$$
choose $cos(2x)=y$
answered Dec 11 '18 at 17:33
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
0
$begingroup$
Hint:
For $$intcos^m(2x)sin^{2n+1}(2x) dx,$$
choose $cos(2x)=y$
answered Dec 11 '18 at 17:33
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
0
0
0
$begingroup$
Hint:
For $$intcos^m(2x)sin^{2n+1}(2x) dx,$$
choose $cos(2x)=y$
answered Dec 11 '18 at 17:33
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
Hint:
For $$intcos^m(2x)sin^{2n+1}(2x) dx,$$
choose $cos(2x)=y$
answered Dec 11 '18 at 17:33
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 11 '18 at 17:33
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 11 '18 at 17:33
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 11 '18 at 17:33
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
add a comment |
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3035478%2fintegration-of-int-0-frac-pi6-cos-32x-sin2x-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
