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













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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




















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