Evaluating $ int frac1{5S_{8} - 9S_{10} + 7S_{12} - 2S_{14}} {rm d}x$, where $S_n = sin^n(x) + cos^n(x)$












5












$begingroup$



Prove that



$$ int frac1{5S_{8} - 9S_{10} + 7S_{12} - 2S_{14}} {rm d}x = 2x - arctan left( frac{tan2x}{2 + tan^22x} right) + C$$



where $S_n = sin^n(x) + cos^n(x)$.




Even differentiating the right doesn't end up with anything close to that monster integrand.



$$ frac{{rm d}}{{rm d}x} left( 2x - arctan left( frac{tan2x}{2 + tan^22x} right) + C right)
\ = 2 + frac{(tan^22x+1)(2tan^22x-4)}{tan^42x+5tan^22x+4}
\ = frac{ 4 } {sin^42x+ 5sin^22xcos^22x + 4cos^42x}
\ = frac1{sin^8x + sin^2xcos^6x +cos^2xsin^6x + cos^8x} $$






integration trigonometry indefinite-integrals trigonometric-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The denominator in your final expression is $S_6$
    $endgroup$
    – Empy2
    Dec 13 '18 at 12:14






  • 2




    $begingroup$
    $S_0=2,S_2=1$ $$S_{m+2}=sin^mx(1-cos^2x)+cos^mx(1-sin^2x)=S_m-sin^2xcos^2xS_{m-2}$$
    $endgroup$
    – lab bhattacharjee
    Dec 13 '18 at 14:38
















5












$begingroup$



Prove that



$$ int frac1{5S_{8} - 9S_{10} + 7S_{12} - 2S_{14}} {rm d}x = 2x - arctan left( frac{tan2x}{2 + tan^22x} right) + C$$



where $S_n = sin^n(x) + cos^n(x)$.




Even differentiating the right doesn't end up with anything close to that monster integrand.



$$ frac{{rm d}}{{rm d}x} left( 2x - arctan left( frac{tan2x}{2 + tan^22x} right) + C right)
\ = 2 + frac{(tan^22x+1)(2tan^22x-4)}{tan^42x+5tan^22x+4}
\ = frac{ 4 } {sin^42x+ 5sin^22xcos^22x + 4cos^42x}
\ = frac1{sin^8x + sin^2xcos^6x +cos^2xsin^6x + cos^8x} $$






integration trigonometry indefinite-integrals trigonometric-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The denominator in your final expression is $S_6$
    $endgroup$
    – Empy2
    Dec 13 '18 at 12:14






  • 2




    $begingroup$
    $S_0=2,S_2=1$ $$S_{m+2}=sin^mx(1-cos^2x)+cos^mx(1-sin^2x)=S_m-sin^2xcos^2xS_{m-2}$$
    $endgroup$
    – lab bhattacharjee
    Dec 13 '18 at 14:38














5












5








5


1



$begingroup$



Prove that



$$ int frac1{5S_{8} - 9S_{10} + 7S_{12} - 2S_{14}} {rm d}x = 2x - arctan left( frac{tan2x}{2 + tan^22x} right) + C$$



where $S_n = sin^n(x) + cos^n(x)$.




Even differentiating the right doesn't end up with anything close to that monster integrand.



$$ frac{{rm d}}{{rm d}x} left( 2x - arctan left( frac{tan2x}{2 + tan^22x} right) + C right)
\ = 2 + frac{(tan^22x+1)(2tan^22x-4)}{tan^42x+5tan^22x+4}
\ = frac{ 4 } {sin^42x+ 5sin^22xcos^22x + 4cos^42x}
\ = frac1{sin^8x + sin^2xcos^6x +cos^2xsin^6x + cos^8x} $$






integration trigonometry indefinite-integrals trigonometric-integrals







share|cite|improve this question











$endgroup$





Prove that



$$ int frac1{5S_{8} - 9S_{10} + 7S_{12} - 2S_{14}} {rm d}x = 2x - arctan left( frac{tan2x}{2 + tan^22x} right) + C$$



where $S_n = sin^n(x) + cos^n(x)$.




Even differentiating the right doesn't end up with anything close to that monster integrand.



$$ frac{{rm d}}{{rm d}x} left( 2x - arctan left( frac{tan2x}{2 + tan^22x} right) + C right)
\ = 2 + frac{(tan^22x+1)(2tan^22x-4)}{tan^42x+5tan^22x+4}
\ = frac{ 4 } {sin^42x+ 5sin^22xcos^22x + 4cos^42x}
\ = frac1{sin^8x + sin^2xcos^6x +cos^2xsin^6x + cos^8x} $$






integration trigonometry indefinite-integrals trigonometric-integrals




integration trigonometry indefinite-integrals trigonometric-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 13:23









Blue

48.6k870156




48.6k870156










asked Dec 13 '18 at 12:08









MintMint

5311417




5311417








  • 1




    $begingroup$
    The denominator in your final expression is $S_6$
    $endgroup$
    – Empy2
    Dec 13 '18 at 12:14






  • 2




    $begingroup$
    $S_0=2,S_2=1$ $$S_{m+2}=sin^mx(1-cos^2x)+cos^mx(1-sin^2x)=S_m-sin^2xcos^2xS_{m-2}$$
    $endgroup$
    – lab bhattacharjee
    Dec 13 '18 at 14:38














  • 1




    $begingroup$
    The denominator in your final expression is $S_6$
    $endgroup$
    – Empy2
    Dec 13 '18 at 12:14






  • 2




    $begingroup$
    $S_0=2,S_2=1$ $$S_{m+2}=sin^mx(1-cos^2x)+cos^mx(1-sin^2x)=S_m-sin^2xcos^2xS_{m-2}$$
    $endgroup$
    – lab bhattacharjee
    Dec 13 '18 at 14:38








1




1




$begingroup$
The denominator in your final expression is $S_6$
$endgroup$
– Empy2
Dec 13 '18 at 12:14




$begingroup$
The denominator in your final expression is $S_6$
$endgroup$
– Empy2
Dec 13 '18 at 12:14




2




2




$begingroup$
$S_0=2,S_2=1$ $$S_{m+2}=sin^mx(1-cos^2x)+cos^mx(1-sin^2x)=S_m-sin^2xcos^2xS_{m-2}$$
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 14:38




$begingroup$
$S_0=2,S_2=1$ $$S_{m+2}=sin^mx(1-cos^2x)+cos^mx(1-sin^2x)=S_m-sin^2xcos^2xS_{m-2}$$
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 14:38










1 Answer
1






active

oldest

votes


















5












$begingroup$

Let $c=cos2x$. Then applying the double angle identities,
$$sin^2x=frac{1-cos2x}2,quadcos^2x=frac{1+cos2x}2,$$
to each occurrence of $sin^nx$ and $cos^nx$, we get
$$begin{cases}
S_8=2^{-3}(1+6c^2+c^4)\
S_{10}=2^{-4}(1+10c^2+5c^4)\
S_{12}=2^{-5}(1+15c^2+15c^4+c^6)\
S_{14}=2^{-6}(1+21c^2+35c^4+7c^6)
end{cases}$$
Combining these expressions to match the denominator of the integrand, we find that it reduces to
$$5S_8-9S_{10}+7S_{12}-2S_{14}=2^{-2}(1+3c^2)$$
so that
$$intfrac{mathrm dx}{5S_8-9S_{10}+7S_{12}-2S_{14}}=intfrac4{1+3cos^22x},mathrm dx=intfrac8{5+3cos4x},mathrm dx$$
The antiderivative is $-arctan(2cot2x)+C$, which I don't think should be too hard to reconcile with the given solution.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    The pure elegance and beauty of this answer is incredible. I commend you with a meaningless upvote.
    $endgroup$
    – clathratus
    Dec 14 '18 at 16:20











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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037919%2fevaluating-int-frac15s-8-9s-10-7s-12-2s-14-rm-dx-where%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









5












$begingroup$

Let $c=cos2x$. Then applying the double angle identities,
$$sin^2x=frac{1-cos2x}2,quadcos^2x=frac{1+cos2x}2,$$
to each occurrence of $sin^nx$ and $cos^nx$, we get
$$begin{cases}
S_8=2^{-3}(1+6c^2+c^4)\
S_{10}=2^{-4}(1+10c^2+5c^4)\
S_{12}=2^{-5}(1+15c^2+15c^4+c^6)\
S_{14}=2^{-6}(1+21c^2+35c^4+7c^6)
end{cases}$$
Combining these expressions to match the denominator of the integrand, we find that it reduces to
$$5S_8-9S_{10}+7S_{12}-2S_{14}=2^{-2}(1+3c^2)$$
so that
$$intfrac{mathrm dx}{5S_8-9S_{10}+7S_{12}-2S_{14}}=intfrac4{1+3cos^22x},mathrm dx=intfrac8{5+3cos4x},mathrm dx$$
The antiderivative is $-arctan(2cot2x)+C$, which I don't think should be too hard to reconcile with the given solution.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    The pure elegance and beauty of this answer is incredible. I commend you with a meaningless upvote.
    $endgroup$
    – clathratus
    Dec 14 '18 at 16:20
















5












$begingroup$

Let $c=cos2x$. Then applying the double angle identities,
$$sin^2x=frac{1-cos2x}2,quadcos^2x=frac{1+cos2x}2,$$
to each occurrence of $sin^nx$ and $cos^nx$, we get
$$begin{cases}
S_8=2^{-3}(1+6c^2+c^4)\
S_{10}=2^{-4}(1+10c^2+5c^4)\
S_{12}=2^{-5}(1+15c^2+15c^4+c^6)\
S_{14}=2^{-6}(1+21c^2+35c^4+7c^6)
end{cases}$$
Combining these expressions to match the denominator of the integrand, we find that it reduces to
$$5S_8-9S_{10}+7S_{12}-2S_{14}=2^{-2}(1+3c^2)$$
so that
$$intfrac{mathrm dx}{5S_8-9S_{10}+7S_{12}-2S_{14}}=intfrac4{1+3cos^22x},mathrm dx=intfrac8{5+3cos4x},mathrm dx$$
The antiderivative is $-arctan(2cot2x)+C$, which I don't think should be too hard to reconcile with the given solution.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    The pure elegance and beauty of this answer is incredible. I commend you with a meaningless upvote.
    $endgroup$
    – clathratus
    Dec 14 '18 at 16:20














5












5








5





$begingroup$

Let $c=cos2x$. Then applying the double angle identities,
$$sin^2x=frac{1-cos2x}2,quadcos^2x=frac{1+cos2x}2,$$
to each occurrence of $sin^nx$ and $cos^nx$, we get
$$begin{cases}
S_8=2^{-3}(1+6c^2+c^4)\
S_{10}=2^{-4}(1+10c^2+5c^4)\
S_{12}=2^{-5}(1+15c^2+15c^4+c^6)\
S_{14}=2^{-6}(1+21c^2+35c^4+7c^6)
end{cases}$$
Combining these expressions to match the denominator of the integrand, we find that it reduces to
$$5S_8-9S_{10}+7S_{12}-2S_{14}=2^{-2}(1+3c^2)$$
so that
$$intfrac{mathrm dx}{5S_8-9S_{10}+7S_{12}-2S_{14}}=intfrac4{1+3cos^22x},mathrm dx=intfrac8{5+3cos4x},mathrm dx$$
The antiderivative is $-arctan(2cot2x)+C$, which I don't think should be too hard to reconcile with the given solution.






share|cite|improve this answer









$endgroup$



Let $c=cos2x$. Then applying the double angle identities,
$$sin^2x=frac{1-cos2x}2,quadcos^2x=frac{1+cos2x}2,$$
to each occurrence of $sin^nx$ and $cos^nx$, we get
$$begin{cases}
S_8=2^{-3}(1+6c^2+c^4)\
S_{10}=2^{-4}(1+10c^2+5c^4)\
S_{12}=2^{-5}(1+15c^2+15c^4+c^6)\
S_{14}=2^{-6}(1+21c^2+35c^4+7c^6)
end{cases}$$
Combining these expressions to match the denominator of the integrand, we find that it reduces to
$$5S_8-9S_{10}+7S_{12}-2S_{14}=2^{-2}(1+3c^2)$$
so that
$$intfrac{mathrm dx}{5S_8-9S_{10}+7S_{12}-2S_{14}}=intfrac4{1+3cos^22x},mathrm dx=intfrac8{5+3cos4x},mathrm dx$$
The antiderivative is $-arctan(2cot2x)+C$, which I don't think should be too hard to reconcile with the given solution.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 17:17









user170231user170231

4,15811429




4,15811429








  • 1




    $begingroup$
    The pure elegance and beauty of this answer is incredible. I commend you with a meaningless upvote.
    $endgroup$
    – clathratus
    Dec 14 '18 at 16:20














  • 1




    $begingroup$
    The pure elegance and beauty of this answer is incredible. I commend you with a meaningless upvote.
    $endgroup$
    – clathratus
    Dec 14 '18 at 16:20








1




1




$begingroup$
The pure elegance and beauty of this answer is incredible. I commend you with a meaningless upvote.
$endgroup$
– clathratus
Dec 14 '18 at 16:20




$begingroup$
The pure elegance and beauty of this answer is incredible. I commend you with a meaningless upvote.
$endgroup$
– clathratus
Dec 14 '18 at 16:20


















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














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037919%2fevaluating-int-frac15s-8-9s-10-7s-12-2s-14-rm-dx-where%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
Read more

In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
Read more

How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

Image
1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
Read more