Evaluating $intfrac{sin x}{sin (x-a)cdot sin (x-b)},mathrm dx$












5












$begingroup$


$$intfrac{sin x}{sin (x-a)cdot sin (x-b)},mathrm dx$$



My Try::



begin{align}
&displaystyle frac{1}{sin (b-a)}intfrac{sin {(x-a)-(x-b)}}{sin (x-a)cdot sin (x-b)}cdotsin x,mathrm dx\ =& frac{1}{sin (b-a)}int left{cot (x-b)-cot (x-a)right}cdot sin x ,mathrm dx
end{align}



Now how can I proceed after that?






calculus indefinite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta.
    $endgroup$
    – Lord_Farin
    Sep 20 '13 at 19:27










  • $begingroup$
    Sorry Lord_Farin , next time i will take care of it.
    $endgroup$
    – juantheron
    Sep 20 '13 at 19:40










  • $begingroup$
    @experimentX Please do not bump questions with such minuscule edits of disputable quality.
    $endgroup$
    – Lord_Farin
    Sep 20 '13 at 20:32






  • 1




    $begingroup$
    You may start from $sin(x)=sin(x-a+a).$
    $endgroup$
    – user64494
    Sep 20 '13 at 20:33










  • $begingroup$
    Step 1: Open those $sin(x-a)$ and $sin(x-b)$ to write everything in terms of $sin(x)$ and $cos(x)$.
    $endgroup$
    – Pp..
    Feb 7 '15 at 1:34
















5












$begingroup$


$$intfrac{sin x}{sin (x-a)cdot sin (x-b)},mathrm dx$$



My Try::



begin{align}
&displaystyle frac{1}{sin (b-a)}intfrac{sin {(x-a)-(x-b)}}{sin (x-a)cdot sin (x-b)}cdotsin x,mathrm dx\ =& frac{1}{sin (b-a)}int left{cot (x-b)-cot (x-a)right}cdot sin x ,mathrm dx
end{align}



Now how can I proceed after that?






calculus indefinite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta.
    $endgroup$
    – Lord_Farin
    Sep 20 '13 at 19:27










  • $begingroup$
    Sorry Lord_Farin , next time i will take care of it.
    $endgroup$
    – juantheron
    Sep 20 '13 at 19:40










  • $begingroup$
    @experimentX Please do not bump questions with such minuscule edits of disputable quality.
    $endgroup$
    – Lord_Farin
    Sep 20 '13 at 20:32






  • 1




    $begingroup$
    You may start from $sin(x)=sin(x-a+a).$
    $endgroup$
    – user64494
    Sep 20 '13 at 20:33










  • $begingroup$
    Step 1: Open those $sin(x-a)$ and $sin(x-b)$ to write everything in terms of $sin(x)$ and $cos(x)$.
    $endgroup$
    – Pp..
    Feb 7 '15 at 1:34














5












5








5


3



$begingroup$


$$intfrac{sin x}{sin (x-a)cdot sin (x-b)},mathrm dx$$



My Try::



begin{align}
&displaystyle frac{1}{sin (b-a)}intfrac{sin {(x-a)-(x-b)}}{sin (x-a)cdot sin (x-b)}cdotsin x,mathrm dx\ =& frac{1}{sin (b-a)}int left{cot (x-b)-cot (x-a)right}cdot sin x ,mathrm dx
end{align}



Now how can I proceed after that?






calculus indefinite-integrals







share|cite|improve this question











$endgroup$




$$intfrac{sin x}{sin (x-a)cdot sin (x-b)},mathrm dx$$



My Try::



begin{align}
&displaystyle frac{1}{sin (b-a)}intfrac{sin {(x-a)-(x-b)}}{sin (x-a)cdot sin (x-b)}cdotsin x,mathrm dx\ =& frac{1}{sin (b-a)}int left{cot (x-b)-cot (x-a)right}cdot sin x ,mathrm dx
end{align}



Now how can I proceed after that?






calculus indefinite-integrals




calculus indefinite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 20 '13 at 19:55









Santosh Linkha

9,64422852




9,64422852










asked Sep 20 '13 at 19:19









juantheronjuantheron

34.4k1150143




34.4k1150143












  • $begingroup$
    Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta.
    $endgroup$
    – Lord_Farin
    Sep 20 '13 at 19:27










  • $begingroup$
    Sorry Lord_Farin , next time i will take care of it.
    $endgroup$
    – juantheron
    Sep 20 '13 at 19:40










  • $begingroup$
    @experimentX Please do not bump questions with such minuscule edits of disputable quality.
    $endgroup$
    – Lord_Farin
    Sep 20 '13 at 20:32






  • 1




    $begingroup$
    You may start from $sin(x)=sin(x-a+a).$
    $endgroup$
    – user64494
    Sep 20 '13 at 20:33










  • $begingroup$
    Step 1: Open those $sin(x-a)$ and $sin(x-b)$ to write everything in terms of $sin(x)$ and $cos(x)$.
    $endgroup$
    – Pp..
    Feb 7 '15 at 1:34


















  • $begingroup$
    Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta.
    $endgroup$
    – Lord_Farin
    Sep 20 '13 at 19:27










  • $begingroup$
    Sorry Lord_Farin , next time i will take care of it.
    $endgroup$
    – juantheron
    Sep 20 '13 at 19:40










  • $begingroup$
    @experimentX Please do not bump questions with such minuscule edits of disputable quality.
    $endgroup$
    – Lord_Farin
    Sep 20 '13 at 20:32






  • 1




    $begingroup$
    You may start from $sin(x)=sin(x-a+a).$
    $endgroup$
    – user64494
    Sep 20 '13 at 20:33










  • $begingroup$
    Step 1: Open those $sin(x-a)$ and $sin(x-b)$ to write everything in terms of $sin(x)$ and $cos(x)$.
    $endgroup$
    – Pp..
    Feb 7 '15 at 1:34
















$begingroup$
Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta.
$endgroup$
– Lord_Farin
Sep 20 '13 at 19:27




$begingroup$
Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta.
$endgroup$
– Lord_Farin
Sep 20 '13 at 19:27












$begingroup$
Sorry Lord_Farin , next time i will take care of it.
$endgroup$
– juantheron
Sep 20 '13 at 19:40




$begingroup$
Sorry Lord_Farin , next time i will take care of it.
$endgroup$
– juantheron
Sep 20 '13 at 19:40












$begingroup$
@experimentX Please do not bump questions with such minuscule edits of disputable quality.
$endgroup$
– Lord_Farin
Sep 20 '13 at 20:32




$begingroup$
@experimentX Please do not bump questions with such minuscule edits of disputable quality.
$endgroup$
– Lord_Farin
Sep 20 '13 at 20:32




1




1




$begingroup$
You may start from $sin(x)=sin(x-a+a).$
$endgroup$
– user64494
Sep 20 '13 at 20:33




$begingroup$
You may start from $sin(x)=sin(x-a+a).$
$endgroup$
– user64494
Sep 20 '13 at 20:33












$begingroup$
Step 1: Open those $sin(x-a)$ and $sin(x-b)$ to write everything in terms of $sin(x)$ and $cos(x)$.
$endgroup$
– Pp..
Feb 7 '15 at 1:34




$begingroup$
Step 1: Open those $sin(x-a)$ and $sin(x-b)$ to write everything in terms of $sin(x)$ and $cos(x)$.
$endgroup$
– Pp..
Feb 7 '15 at 1:34










1 Answer
1






active

oldest

votes


















3












$begingroup$

Wrong answer for now.



First let's rewrite:
$$frac{1}{sin(x-a)sin(x-b)}=frac{p}{sin(x-a)}+frac{q}{sin(x-b)} quad(1)$$
Now multiplying $(1)$ by $sin(x-a)$ followed by letting $x=a$ yields:
$$left(frac{1}{sin(x-b)}=p+qunderset{rightarrow 0}{frac{sin(x-a)}{sin(x-b)}}right)bigg|_{x=a}Rightarrow p=frac{1}{sin(a-b)}$$
Similarly to obtain q, multiply $(1)$ by $sin(x-b)$ and plugg $x=b$ gives:
$$left(frac{1}{sin(x-a)}=punderset{rightarrow 0}{frac{sin(x-b)}{sin(x-a)}}+qright)bigg|_{x=b}Rightarrow q=frac{1}{sin(b-a)}=-p$$
Back to the original integral, we have:
$$I=intfrac{sin x}{sin (x-a)cdot sin (x-b)}dx=frac{1}{sin(a-b)}left(intfrac{sin x}{sin (x-a)}dx-intfrac{sin x}{sin(x-b)}dxright)$$
Now, it's enough to solve only one of these integrals.
$$intfrac{sin x}{sin(x-t)}dx=intfrac{sin(x-t+t)}{sin (x-t)}dx=cos tintfrac{sin(x-t)}{sin(x-t)}dx+sin tintfrac{cos(x-t)}{sin(x-t)}dx=$$
$$=cos t cdot x +sin t cdot ln|sin(x-t)|+c$$
Thus replacing $t$ by $a$ and $b$ we get the answer to be:
$$I=frac{1}{sin(a-b)}left((cos a-cos b)x+sin a ln|sin(x-a)|-sin b ln|sin(x-b)|right)+C$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Unfortunately,$$frac{1}{sinleft(a-bright)}left(frac{1}{sinleft(x-aright)}-frac{1}{sinleft(x-bright)}right)=frac{sinleft(x-bright)-sinleft(x-aright)}{sinleft(a-bright)sinleft(x-aright)sinleft(x-bright)}=frac{cosleft(x-frac{a+b}{2}right)}{cosfrac{a-b}{2}sinleft(x-aright)sinleft(x-bright)}notequivfrac{1}{sinleft(x-aright)sinleft(x-bright)}.$$
    $endgroup$
    – J.G.
    Dec 20 '18 at 12:08










  • $begingroup$
    This is quite weird, I don't see where is my error.
    $endgroup$
    – Zacky
    Dec 20 '18 at 12:14








  • 1




    $begingroup$
    Your error is assuming constant coefficients exist that get the job done in the first place.
    $endgroup$
    – J.G.
    Dec 20 '18 at 12:48













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%2f499849%2fevaluating-int-frac-sin-x-sin-x-a-cdot-sin-x-b-mathrm-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









3












$begingroup$

Wrong answer for now.



First let's rewrite:
$$frac{1}{sin(x-a)sin(x-b)}=frac{p}{sin(x-a)}+frac{q}{sin(x-b)} quad(1)$$
Now multiplying $(1)$ by $sin(x-a)$ followed by letting $x=a$ yields:
$$left(frac{1}{sin(x-b)}=p+qunderset{rightarrow 0}{frac{sin(x-a)}{sin(x-b)}}right)bigg|_{x=a}Rightarrow p=frac{1}{sin(a-b)}$$
Similarly to obtain q, multiply $(1)$ by $sin(x-b)$ and plugg $x=b$ gives:
$$left(frac{1}{sin(x-a)}=punderset{rightarrow 0}{frac{sin(x-b)}{sin(x-a)}}+qright)bigg|_{x=b}Rightarrow q=frac{1}{sin(b-a)}=-p$$
Back to the original integral, we have:
$$I=intfrac{sin x}{sin (x-a)cdot sin (x-b)}dx=frac{1}{sin(a-b)}left(intfrac{sin x}{sin (x-a)}dx-intfrac{sin x}{sin(x-b)}dxright)$$
Now, it's enough to solve only one of these integrals.
$$intfrac{sin x}{sin(x-t)}dx=intfrac{sin(x-t+t)}{sin (x-t)}dx=cos tintfrac{sin(x-t)}{sin(x-t)}dx+sin tintfrac{cos(x-t)}{sin(x-t)}dx=$$
$$=cos t cdot x +sin t cdot ln|sin(x-t)|+c$$
Thus replacing $t$ by $a$ and $b$ we get the answer to be:
$$I=frac{1}{sin(a-b)}left((cos a-cos b)x+sin a ln|sin(x-a)|-sin b ln|sin(x-b)|right)+C$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Unfortunately,$$frac{1}{sinleft(a-bright)}left(frac{1}{sinleft(x-aright)}-frac{1}{sinleft(x-bright)}right)=frac{sinleft(x-bright)-sinleft(x-aright)}{sinleft(a-bright)sinleft(x-aright)sinleft(x-bright)}=frac{cosleft(x-frac{a+b}{2}right)}{cosfrac{a-b}{2}sinleft(x-aright)sinleft(x-bright)}notequivfrac{1}{sinleft(x-aright)sinleft(x-bright)}.$$
    $endgroup$
    – J.G.
    Dec 20 '18 at 12:08










  • $begingroup$
    This is quite weird, I don't see where is my error.
    $endgroup$
    – Zacky
    Dec 20 '18 at 12:14








  • 1




    $begingroup$
    Your error is assuming constant coefficients exist that get the job done in the first place.
    $endgroup$
    – J.G.
    Dec 20 '18 at 12:48


















3












$begingroup$

Wrong answer for now.



First let's rewrite:
$$frac{1}{sin(x-a)sin(x-b)}=frac{p}{sin(x-a)}+frac{q}{sin(x-b)} quad(1)$$
Now multiplying $(1)$ by $sin(x-a)$ followed by letting $x=a$ yields:
$$left(frac{1}{sin(x-b)}=p+qunderset{rightarrow 0}{frac{sin(x-a)}{sin(x-b)}}right)bigg|_{x=a}Rightarrow p=frac{1}{sin(a-b)}$$
Similarly to obtain q, multiply $(1)$ by $sin(x-b)$ and plugg $x=b$ gives:
$$left(frac{1}{sin(x-a)}=punderset{rightarrow 0}{frac{sin(x-b)}{sin(x-a)}}+qright)bigg|_{x=b}Rightarrow q=frac{1}{sin(b-a)}=-p$$
Back to the original integral, we have:
$$I=intfrac{sin x}{sin (x-a)cdot sin (x-b)}dx=frac{1}{sin(a-b)}left(intfrac{sin x}{sin (x-a)}dx-intfrac{sin x}{sin(x-b)}dxright)$$
Now, it's enough to solve only one of these integrals.
$$intfrac{sin x}{sin(x-t)}dx=intfrac{sin(x-t+t)}{sin (x-t)}dx=cos tintfrac{sin(x-t)}{sin(x-t)}dx+sin tintfrac{cos(x-t)}{sin(x-t)}dx=$$
$$=cos t cdot x +sin t cdot ln|sin(x-t)|+c$$
Thus replacing $t$ by $a$ and $b$ we get the answer to be:
$$I=frac{1}{sin(a-b)}left((cos a-cos b)x+sin a ln|sin(x-a)|-sin b ln|sin(x-b)|right)+C$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Unfortunately,$$frac{1}{sinleft(a-bright)}left(frac{1}{sinleft(x-aright)}-frac{1}{sinleft(x-bright)}right)=frac{sinleft(x-bright)-sinleft(x-aright)}{sinleft(a-bright)sinleft(x-aright)sinleft(x-bright)}=frac{cosleft(x-frac{a+b}{2}right)}{cosfrac{a-b}{2}sinleft(x-aright)sinleft(x-bright)}notequivfrac{1}{sinleft(x-aright)sinleft(x-bright)}.$$
    $endgroup$
    – J.G.
    Dec 20 '18 at 12:08










  • $begingroup$
    This is quite weird, I don't see where is my error.
    $endgroup$
    – Zacky
    Dec 20 '18 at 12:14








  • 1




    $begingroup$
    Your error is assuming constant coefficients exist that get the job done in the first place.
    $endgroup$
    – J.G.
    Dec 20 '18 at 12:48
















3












3








3





$begingroup$

Wrong answer for now.



First let's rewrite:
$$frac{1}{sin(x-a)sin(x-b)}=frac{p}{sin(x-a)}+frac{q}{sin(x-b)} quad(1)$$
Now multiplying $(1)$ by $sin(x-a)$ followed by letting $x=a$ yields:
$$left(frac{1}{sin(x-b)}=p+qunderset{rightarrow 0}{frac{sin(x-a)}{sin(x-b)}}right)bigg|_{x=a}Rightarrow p=frac{1}{sin(a-b)}$$
Similarly to obtain q, multiply $(1)$ by $sin(x-b)$ and plugg $x=b$ gives:
$$left(frac{1}{sin(x-a)}=punderset{rightarrow 0}{frac{sin(x-b)}{sin(x-a)}}+qright)bigg|_{x=b}Rightarrow q=frac{1}{sin(b-a)}=-p$$
Back to the original integral, we have:
$$I=intfrac{sin x}{sin (x-a)cdot sin (x-b)}dx=frac{1}{sin(a-b)}left(intfrac{sin x}{sin (x-a)}dx-intfrac{sin x}{sin(x-b)}dxright)$$
Now, it's enough to solve only one of these integrals.
$$intfrac{sin x}{sin(x-t)}dx=intfrac{sin(x-t+t)}{sin (x-t)}dx=cos tintfrac{sin(x-t)}{sin(x-t)}dx+sin tintfrac{cos(x-t)}{sin(x-t)}dx=$$
$$=cos t cdot x +sin t cdot ln|sin(x-t)|+c$$
Thus replacing $t$ by $a$ and $b$ we get the answer to be:
$$I=frac{1}{sin(a-b)}left((cos a-cos b)x+sin a ln|sin(x-a)|-sin b ln|sin(x-b)|right)+C$$






share|cite|improve this answer











$endgroup$



Wrong answer for now.



First let's rewrite:
$$frac{1}{sin(x-a)sin(x-b)}=frac{p}{sin(x-a)}+frac{q}{sin(x-b)} quad(1)$$
Now multiplying $(1)$ by $sin(x-a)$ followed by letting $x=a$ yields:
$$left(frac{1}{sin(x-b)}=p+qunderset{rightarrow 0}{frac{sin(x-a)}{sin(x-b)}}right)bigg|_{x=a}Rightarrow p=frac{1}{sin(a-b)}$$
Similarly to obtain q, multiply $(1)$ by $sin(x-b)$ and plugg $x=b$ gives:
$$left(frac{1}{sin(x-a)}=punderset{rightarrow 0}{frac{sin(x-b)}{sin(x-a)}}+qright)bigg|_{x=b}Rightarrow q=frac{1}{sin(b-a)}=-p$$
Back to the original integral, we have:
$$I=intfrac{sin x}{sin (x-a)cdot sin (x-b)}dx=frac{1}{sin(a-b)}left(intfrac{sin x}{sin (x-a)}dx-intfrac{sin x}{sin(x-b)}dxright)$$
Now, it's enough to solve only one of these integrals.
$$intfrac{sin x}{sin(x-t)}dx=intfrac{sin(x-t+t)}{sin (x-t)}dx=cos tintfrac{sin(x-t)}{sin(x-t)}dx+sin tintfrac{cos(x-t)}{sin(x-t)}dx=$$
$$=cos t cdot x +sin t cdot ln|sin(x-t)|+c$$
Thus replacing $t$ by $a$ and $b$ we get the answer to be:
$$I=frac{1}{sin(a-b)}left((cos a-cos b)x+sin a ln|sin(x-a)|-sin b ln|sin(x-b)|right)+C$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 20 '18 at 13:14

























answered Dec 20 '18 at 11:51









ZackyZacky

7,89511061




7,89511061








  • 1




    $begingroup$
    Unfortunately,$$frac{1}{sinleft(a-bright)}left(frac{1}{sinleft(x-aright)}-frac{1}{sinleft(x-bright)}right)=frac{sinleft(x-bright)-sinleft(x-aright)}{sinleft(a-bright)sinleft(x-aright)sinleft(x-bright)}=frac{cosleft(x-frac{a+b}{2}right)}{cosfrac{a-b}{2}sinleft(x-aright)sinleft(x-bright)}notequivfrac{1}{sinleft(x-aright)sinleft(x-bright)}.$$
    $endgroup$
    – J.G.
    Dec 20 '18 at 12:08










  • $begingroup$
    This is quite weird, I don't see where is my error.
    $endgroup$
    – Zacky
    Dec 20 '18 at 12:14








  • 1




    $begingroup$
    Your error is assuming constant coefficients exist that get the job done in the first place.
    $endgroup$
    – J.G.
    Dec 20 '18 at 12:48
















  • 1




    $begingroup$
    Unfortunately,$$frac{1}{sinleft(a-bright)}left(frac{1}{sinleft(x-aright)}-frac{1}{sinleft(x-bright)}right)=frac{sinleft(x-bright)-sinleft(x-aright)}{sinleft(a-bright)sinleft(x-aright)sinleft(x-bright)}=frac{cosleft(x-frac{a+b}{2}right)}{cosfrac{a-b}{2}sinleft(x-aright)sinleft(x-bright)}notequivfrac{1}{sinleft(x-aright)sinleft(x-bright)}.$$
    $endgroup$
    – J.G.
    Dec 20 '18 at 12:08










  • $begingroup$
    This is quite weird, I don't see where is my error.
    $endgroup$
    – Zacky
    Dec 20 '18 at 12:14








  • 1




    $begingroup$
    Your error is assuming constant coefficients exist that get the job done in the first place.
    $endgroup$
    – J.G.
    Dec 20 '18 at 12:48










1




1




$begingroup$
Unfortunately,$$frac{1}{sinleft(a-bright)}left(frac{1}{sinleft(x-aright)}-frac{1}{sinleft(x-bright)}right)=frac{sinleft(x-bright)-sinleft(x-aright)}{sinleft(a-bright)sinleft(x-aright)sinleft(x-bright)}=frac{cosleft(x-frac{a+b}{2}right)}{cosfrac{a-b}{2}sinleft(x-aright)sinleft(x-bright)}notequivfrac{1}{sinleft(x-aright)sinleft(x-bright)}.$$
$endgroup$
– J.G.
Dec 20 '18 at 12:08




$begingroup$
Unfortunately,$$frac{1}{sinleft(a-bright)}left(frac{1}{sinleft(x-aright)}-frac{1}{sinleft(x-bright)}right)=frac{sinleft(x-bright)-sinleft(x-aright)}{sinleft(a-bright)sinleft(x-aright)sinleft(x-bright)}=frac{cosleft(x-frac{a+b}{2}right)}{cosfrac{a-b}{2}sinleft(x-aright)sinleft(x-bright)}notequivfrac{1}{sinleft(x-aright)sinleft(x-bright)}.$$
$endgroup$
– J.G.
Dec 20 '18 at 12:08












$begingroup$
This is quite weird, I don't see where is my error.
$endgroup$
– Zacky
Dec 20 '18 at 12:14






$begingroup$
This is quite weird, I don't see where is my error.
$endgroup$
– Zacky
Dec 20 '18 at 12:14






1




1




$begingroup$
Your error is assuming constant coefficients exist that get the job done in the first place.
$endgroup$
– J.G.
Dec 20 '18 at 12:48






$begingroup$
Your error is assuming constant coefficients exist that get the job done in the first place.
$endgroup$
– J.G.
Dec 20 '18 at 12:48




















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%2f499849%2fevaluating-int-frac-sin-x-sin-x-a-cdot-sin-x-b-mathrm-dx%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

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Verónica Boquete

Image
Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more