How to compute $int_0^pi {sin xsin nxover 1-2acos x+a^2} dx$Compute $int_0^pifrac{cos nx}{a^2-2abcos x+b^2},...

Output with the same length always

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How to compute $int_0^pi {sin xsin nxover 1-2acos x+a^2} dx$


Compute $int_0^pifrac{cos nx}{a^2-2abcos x+b^2}, dx$Proof that $int_0^{2pi}sin nx,dx=int_0^{2pi}cos nx,dx=0$How to find $int_0^{pi}frac{sin ntheta}{costheta-cosalpha}dtheta$Integrating $displaystyleint_0^{pi/2} {sin^2x over 1 + sin xcos x}dx$Prove that $int_0^frac{pi}{2}cos^mx sin^mxdx=2^{-m}int_0^frac{pi}{2}cos^mxdx$Evaluate $int_0^{frac{pi}{2}}frac{sin xcos x}{sin^4x+cos^4x}dx$Tricky real integral: $int_0^{2 pi} e^{cos(2 t)} cos(sin(2 t)) =2pi$$I =int_0^{2pi} ((e^{|sin x|} cos x)/(1+e^{tan x})) ,dx$






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$$int_0^pi {sin xsin nxover 1-2acos x+a^2} dx$$



With some bit of tinkering with Desmos, I've got to know that the answer is ${pi over 2} a^{n-1}$.



But can you help prove that?




Sorry for the typo. I meant $sin nx$ instead of $cos nx$.






integration definite-integrals







share|cite|improve this question











$endgroup$










  • 4




    $begingroup$
    So for $n=1$, you expect an answer independent of $a$...?
    $endgroup$
    – StackTD
    2 days ago








  • 2




    $begingroup$
    I think $a^{2}$ times LHS tends to $0$ for any $n$ but $a^{2}$ times RHS tends to $infty$ as $ a to infty$. The result is false.
    $endgroup$
    – Kavi Rama Murthy
    2 days ago










  • $begingroup$
    desmos.com/calculator/zowwz6h8mo Check this out.
    $endgroup$
    – Atom
    2 days ago










  • $begingroup$
    @Atom Please mention that your result is valid when $a^2<1$. You may see my solution.
    $endgroup$
    – Dr Zafar Ahmed DSc
    2 days ago










  • $begingroup$
    @DrZafarAhmedDSc Yes!
    $endgroup$
    – Atom
    2 days ago


















3












$begingroup$


$$int_0^pi {sin xsin nxover 1-2acos x+a^2} dx$$



With some bit of tinkering with Desmos, I've got to know that the answer is ${pi over 2} a^{n-1}$.



But can you help prove that?




Sorry for the typo. I meant $sin nx$ instead of $cos nx$.






integration definite-integrals







share|cite|improve this question











$endgroup$










  • 4




    $begingroup$
    So for $n=1$, you expect an answer independent of $a$...?
    $endgroup$
    – StackTD
    2 days ago








  • 2




    $begingroup$
    I think $a^{2}$ times LHS tends to $0$ for any $n$ but $a^{2}$ times RHS tends to $infty$ as $ a to infty$. The result is false.
    $endgroup$
    – Kavi Rama Murthy
    2 days ago










  • $begingroup$
    desmos.com/calculator/zowwz6h8mo Check this out.
    $endgroup$
    – Atom
    2 days ago










  • $begingroup$
    @Atom Please mention that your result is valid when $a^2<1$. You may see my solution.
    $endgroup$
    – Dr Zafar Ahmed DSc
    2 days ago










  • $begingroup$
    @DrZafarAhmedDSc Yes!
    $endgroup$
    – Atom
    2 days ago














3












3








3


3



$begingroup$


$$int_0^pi {sin xsin nxover 1-2acos x+a^2} dx$$



With some bit of tinkering with Desmos, I've got to know that the answer is ${pi over 2} a^{n-1}$.



But can you help prove that?




Sorry for the typo. I meant $sin nx$ instead of $cos nx$.






integration definite-integrals







share|cite|improve this question











$endgroup$




$$int_0^pi {sin xsin nxover 1-2acos x+a^2} dx$$



With some bit of tinkering with Desmos, I've got to know that the answer is ${pi over 2} a^{n-1}$.



But can you help prove that?




Sorry for the typo. I meant $sin nx$ instead of $cos nx$.






integration definite-integrals




integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Asaf Karagila♦

315k35 gold badges454 silver badges788 bronze badges




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asked 2 days ago









AtomAtom

4581 silver badge9 bronze badges




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  • 4




    $begingroup$
    So for $n=1$, you expect an answer independent of $a$...?
    $endgroup$
    – StackTD
    2 days ago








  • 2




    $begingroup$
    I think $a^{2}$ times LHS tends to $0$ for any $n$ but $a^{2}$ times RHS tends to $infty$ as $ a to infty$. The result is false.
    $endgroup$
    – Kavi Rama Murthy
    2 days ago










  • $begingroup$
    desmos.com/calculator/zowwz6h8mo Check this out.
    $endgroup$
    – Atom
    2 days ago










  • $begingroup$
    @Atom Please mention that your result is valid when $a^2<1$. You may see my solution.
    $endgroup$
    – Dr Zafar Ahmed DSc
    2 days ago










  • $begingroup$
    @DrZafarAhmedDSc Yes!
    $endgroup$
    – Atom
    2 days ago














  • 4




    $begingroup$
    So for $n=1$, you expect an answer independent of $a$...?
    $endgroup$
    – StackTD
    2 days ago








  • 2




    $begingroup$
    I think $a^{2}$ times LHS tends to $0$ for any $n$ but $a^{2}$ times RHS tends to $infty$ as $ a to infty$. The result is false.
    $endgroup$
    – Kavi Rama Murthy
    2 days ago










  • $begingroup$
    desmos.com/calculator/zowwz6h8mo Check this out.
    $endgroup$
    – Atom
    2 days ago










  • $begingroup$
    @Atom Please mention that your result is valid when $a^2<1$. You may see my solution.
    $endgroup$
    – Dr Zafar Ahmed DSc
    2 days ago










  • $begingroup$
    @DrZafarAhmedDSc Yes!
    $endgroup$
    – Atom
    2 days ago








4




4




$begingroup$
So for $n=1$, you expect an answer independent of $a$...?
$endgroup$
– StackTD
2 days ago






$begingroup$
So for $n=1$, you expect an answer independent of $a$...?
$endgroup$
– StackTD
2 days ago






2




2




$begingroup$
I think $a^{2}$ times LHS tends to $0$ for any $n$ but $a^{2}$ times RHS tends to $infty$ as $ a to infty$. The result is false.
$endgroup$
– Kavi Rama Murthy
2 days ago




$begingroup$
I think $a^{2}$ times LHS tends to $0$ for any $n$ but $a^{2}$ times RHS tends to $infty$ as $ a to infty$. The result is false.
$endgroup$
– Kavi Rama Murthy
2 days ago












$begingroup$
desmos.com/calculator/zowwz6h8mo Check this out.
$endgroup$
– Atom
2 days ago




$begingroup$
desmos.com/calculator/zowwz6h8mo Check this out.
$endgroup$
– Atom
2 days ago












$begingroup$
@Atom Please mention that your result is valid when $a^2<1$. You may see my solution.
$endgroup$
– Dr Zafar Ahmed DSc
2 days ago




$begingroup$
@Atom Please mention that your result is valid when $a^2<1$. You may see my solution.
$endgroup$
– Dr Zafar Ahmed DSc
2 days ago












$begingroup$
@DrZafarAhmedDSc Yes!
$endgroup$
– Atom
2 days ago




$begingroup$
@DrZafarAhmedDSc Yes!
$endgroup$
– Atom
2 days ago










3 Answers
3






active

oldest

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$begingroup$

We can do something more general using a result proven here, namely:
$$int_0^pifrac{cos(mx)}{a^2-2abcos x+b^2}dx=frac{pi}{a^2-b^2}left(frac{b}{a}right)^m$$



$$int_0^pi frac{sin(kx)sin(n x)}{a^2-2abcos x+b^2}dx=frac12int_0^pi frac{cos((k-n)x)-cos((k+n)x)}{a^2-2abcos x+b^2}dx$$
$$=frac{pi}{2(a^2-b^2)}left(left(frac{b}{a}right)^{n-k}-left(frac{b}{a}right)^{n+k}right),quad n>k; a>b>0.$$
Just swap $(n,k)$ and $(a,b)$ for other conditions.






share|cite|improve this answer











$endgroup$















  • $begingroup$
    Should the exponent near the end be $n-k$ instead of $k-n$ ?
    $endgroup$
    – Empy2
    2 days ago










  • $begingroup$
    Is it fine now?
    $endgroup$
    – Zacky
    2 days ago






  • 1




    $begingroup$
    I think it is right.
    $endgroup$
    – Empy2
    2 days ago



















2












$begingroup$

Partial Answer



This function is even so we can write$$int_0^pi {sin xsin nxover 1-2acos x+a^2} dx={1over 2}int_{-pi}^pi {sin xsin nxover 1-2acos x+a^2} dx$$by defining $z=e^{ix}$ we have$$int_{-pi}^pi {sin xsin nxover 1-2acos x+a^2} dx{=oint_{|z|=1} {{1over 2i}(z-z^{-1}){1over 2i}(z^{n}-z^{-n})over 1-aleft(z+{1over z}right)+a^2} {dzover iz}\=-{1over 4}oint_{|z|=1} {(z-z^{-1})(z^{n}-z^{-n})over 1-aleft(z+{1over z}right)+a^2} {dzover iz}\=-{1over 4}oint_{|z|=1} {(z^2-1)(z^{2n}-1)over z^n((1+a^2)z-aleft(z^2+1right))} {dzover iz}\={1over 4ai}oint_{|z|=1} {(z^2-1)(z^{2n}-1)over z^{n+1}(z-a)left(z-{1over a}right)} {dz}}$$with the singularities of ${(z^2-1)(z^{2n}-1)over z^{n+1}(z-a)left(z-{1over a}right)}$ falling in $z={1over a},0,a$ when $ane 1$. For $r=1$, the only singularity exists in $z=0$.






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    1












    $begingroup$

    $$I=int_{0}^{pi} frac {sin x sin n x}{1-2a cos x+a^2} dx ~~~~(1)$$ Integrate it by parts taking $sin x$ as the first function, we get
    $$I=-frac{n}{2a}int_{0}^{pi} cos n x ln[1-2a cos x +a^2] dx ~~~~(2).$$
    Let $y=cos x+i sin x$, then $$f(x)=ln(1-2acos x+a^2)=ln(1-ay)+ln(1-a/y).$$
    If $a^2<1,$ then $$f(x)=-2[a cos x + frac{a^2cos 2 x}{2}+frac{a^3 cos 3 x}{3}+...]~~~~(3)$$



    Using (3) in (2) and using the property that $$int_{0}^{pi} cos m x cos n x dx=frac{pi}{2}delta_{m,n}$$ we get
    $$I=frac{n}{a} frac{a^n}{n} frac{pi}{2}=frac{pi a^{n-1}}{2},~~~ a^2 <1.$$






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      We can do something more general using a result proven here, namely:
      $$int_0^pifrac{cos(mx)}{a^2-2abcos x+b^2}dx=frac{pi}{a^2-b^2}left(frac{b}{a}right)^m$$



      $$int_0^pi frac{sin(kx)sin(n x)}{a^2-2abcos x+b^2}dx=frac12int_0^pi frac{cos((k-n)x)-cos((k+n)x)}{a^2-2abcos x+b^2}dx$$
      $$=frac{pi}{2(a^2-b^2)}left(left(frac{b}{a}right)^{n-k}-left(frac{b}{a}right)^{n+k}right),quad n>k; a>b>0.$$
      Just swap $(n,k)$ and $(a,b)$ for other conditions.






      share|cite|improve this answer











      $endgroup$















      • $begingroup$
        Should the exponent near the end be $n-k$ instead of $k-n$ ?
        $endgroup$
        – Empy2
        2 days ago










      • $begingroup$
        Is it fine now?
        $endgroup$
        – Zacky
        2 days ago






      • 1




        $begingroup$
        I think it is right.
        $endgroup$
        – Empy2
        2 days ago
















      4












      $begingroup$

      We can do something more general using a result proven here, namely:
      $$int_0^pifrac{cos(mx)}{a^2-2abcos x+b^2}dx=frac{pi}{a^2-b^2}left(frac{b}{a}right)^m$$



      $$int_0^pi frac{sin(kx)sin(n x)}{a^2-2abcos x+b^2}dx=frac12int_0^pi frac{cos((k-n)x)-cos((k+n)x)}{a^2-2abcos x+b^2}dx$$
      $$=frac{pi}{2(a^2-b^2)}left(left(frac{b}{a}right)^{n-k}-left(frac{b}{a}right)^{n+k}right),quad n>k; a>b>0.$$
      Just swap $(n,k)$ and $(a,b)$ for other conditions.






      share|cite|improve this answer











      $endgroup$















      • $begingroup$
        Should the exponent near the end be $n-k$ instead of $k-n$ ?
        $endgroup$
        – Empy2
        2 days ago










      • $begingroup$
        Is it fine now?
        $endgroup$
        – Zacky
        2 days ago






      • 1




        $begingroup$
        I think it is right.
        $endgroup$
        – Empy2
        2 days ago














      4












      4








      4





      $begingroup$

      We can do something more general using a result proven here, namely:
      $$int_0^pifrac{cos(mx)}{a^2-2abcos x+b^2}dx=frac{pi}{a^2-b^2}left(frac{b}{a}right)^m$$



      $$int_0^pi frac{sin(kx)sin(n x)}{a^2-2abcos x+b^2}dx=frac12int_0^pi frac{cos((k-n)x)-cos((k+n)x)}{a^2-2abcos x+b^2}dx$$
      $$=frac{pi}{2(a^2-b^2)}left(left(frac{b}{a}right)^{n-k}-left(frac{b}{a}right)^{n+k}right),quad n>k; a>b>0.$$
      Just swap $(n,k)$ and $(a,b)$ for other conditions.






      share|cite|improve this answer











      $endgroup$



      We can do something more general using a result proven here, namely:
      $$int_0^pifrac{cos(mx)}{a^2-2abcos x+b^2}dx=frac{pi}{a^2-b^2}left(frac{b}{a}right)^m$$



      $$int_0^pi frac{sin(kx)sin(n x)}{a^2-2abcos x+b^2}dx=frac12int_0^pi frac{cos((k-n)x)-cos((k+n)x)}{a^2-2abcos x+b^2}dx$$
      $$=frac{pi}{2(a^2-b^2)}left(left(frac{b}{a}right)^{n-k}-left(frac{b}{a}right)^{n+k}right),quad n>k; a>b>0.$$
      Just swap $(n,k)$ and $(a,b)$ for other conditions.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 2 days ago

























      answered 2 days ago









      ZackyZacky

      13.2k1 gold badge21 silver badges85 bronze badges




      13.2k1 gold badge21 silver badges85 bronze badges















      • $begingroup$
        Should the exponent near the end be $n-k$ instead of $k-n$ ?
        $endgroup$
        – Empy2
        2 days ago










      • $begingroup$
        Is it fine now?
        $endgroup$
        – Zacky
        2 days ago






      • 1




        $begingroup$
        I think it is right.
        $endgroup$
        – Empy2
        2 days ago


















      • $begingroup$
        Should the exponent near the end be $n-k$ instead of $k-n$ ?
        $endgroup$
        – Empy2
        2 days ago










      • $begingroup$
        Is it fine now?
        $endgroup$
        – Zacky
        2 days ago






      • 1




        $begingroup$
        I think it is right.
        $endgroup$
        – Empy2
        2 days ago
















      $begingroup$
      Should the exponent near the end be $n-k$ instead of $k-n$ ?
      $endgroup$
      – Empy2
      2 days ago




      $begingroup$
      Should the exponent near the end be $n-k$ instead of $k-n$ ?
      $endgroup$
      – Empy2
      2 days ago












      $begingroup$
      Is it fine now?
      $endgroup$
      – Zacky
      2 days ago




      $begingroup$
      Is it fine now?
      $endgroup$
      – Zacky
      2 days ago




      1




      1




      $begingroup$
      I think it is right.
      $endgroup$
      – Empy2
      2 days ago




      $begingroup$
      I think it is right.
      $endgroup$
      – Empy2
      2 days ago













      2












      $begingroup$

      Partial Answer



      This function is even so we can write$$int_0^pi {sin xsin nxover 1-2acos x+a^2} dx={1over 2}int_{-pi}^pi {sin xsin nxover 1-2acos x+a^2} dx$$by defining $z=e^{ix}$ we have$$int_{-pi}^pi {sin xsin nxover 1-2acos x+a^2} dx{=oint_{|z|=1} {{1over 2i}(z-z^{-1}){1over 2i}(z^{n}-z^{-n})over 1-aleft(z+{1over z}right)+a^2} {dzover iz}\=-{1over 4}oint_{|z|=1} {(z-z^{-1})(z^{n}-z^{-n})over 1-aleft(z+{1over z}right)+a^2} {dzover iz}\=-{1over 4}oint_{|z|=1} {(z^2-1)(z^{2n}-1)over z^n((1+a^2)z-aleft(z^2+1right))} {dzover iz}\={1over 4ai}oint_{|z|=1} {(z^2-1)(z^{2n}-1)over z^{n+1}(z-a)left(z-{1over a}right)} {dz}}$$with the singularities of ${(z^2-1)(z^{2n}-1)over z^{n+1}(z-a)left(z-{1over a}right)}$ falling in $z={1over a},0,a$ when $ane 1$. For $r=1$, the only singularity exists in $z=0$.






      share|cite|improve this answer









      $endgroup$




















        2












        $begingroup$

        Partial Answer



        This function is even so we can write$$int_0^pi {sin xsin nxover 1-2acos x+a^2} dx={1over 2}int_{-pi}^pi {sin xsin nxover 1-2acos x+a^2} dx$$by defining $z=e^{ix}$ we have$$int_{-pi}^pi {sin xsin nxover 1-2acos x+a^2} dx{=oint_{|z|=1} {{1over 2i}(z-z^{-1}){1over 2i}(z^{n}-z^{-n})over 1-aleft(z+{1over z}right)+a^2} {dzover iz}\=-{1over 4}oint_{|z|=1} {(z-z^{-1})(z^{n}-z^{-n})over 1-aleft(z+{1over z}right)+a^2} {dzover iz}\=-{1over 4}oint_{|z|=1} {(z^2-1)(z^{2n}-1)over z^n((1+a^2)z-aleft(z^2+1right))} {dzover iz}\={1over 4ai}oint_{|z|=1} {(z^2-1)(z^{2n}-1)over z^{n+1}(z-a)left(z-{1over a}right)} {dz}}$$with the singularities of ${(z^2-1)(z^{2n}-1)over z^{n+1}(z-a)left(z-{1over a}right)}$ falling in $z={1over a},0,a$ when $ane 1$. For $r=1$, the only singularity exists in $z=0$.






        share|cite|improve this answer









        $endgroup$


















          2












          2








          2





          $begingroup$

          Partial Answer



          This function is even so we can write$$int_0^pi {sin xsin nxover 1-2acos x+a^2} dx={1over 2}int_{-pi}^pi {sin xsin nxover 1-2acos x+a^2} dx$$by defining $z=e^{ix}$ we have$$int_{-pi}^pi {sin xsin nxover 1-2acos x+a^2} dx{=oint_{|z|=1} {{1over 2i}(z-z^{-1}){1over 2i}(z^{n}-z^{-n})over 1-aleft(z+{1over z}right)+a^2} {dzover iz}\=-{1over 4}oint_{|z|=1} {(z-z^{-1})(z^{n}-z^{-n})over 1-aleft(z+{1over z}right)+a^2} {dzover iz}\=-{1over 4}oint_{|z|=1} {(z^2-1)(z^{2n}-1)over z^n((1+a^2)z-aleft(z^2+1right))} {dzover iz}\={1over 4ai}oint_{|z|=1} {(z^2-1)(z^{2n}-1)over z^{n+1}(z-a)left(z-{1over a}right)} {dz}}$$with the singularities of ${(z^2-1)(z^{2n}-1)over z^{n+1}(z-a)left(z-{1over a}right)}$ falling in $z={1over a},0,a$ when $ane 1$. For $r=1$, the only singularity exists in $z=0$.






          share|cite|improve this answer









          $endgroup$



          Partial Answer



          This function is even so we can write$$int_0^pi {sin xsin nxover 1-2acos x+a^2} dx={1over 2}int_{-pi}^pi {sin xsin nxover 1-2acos x+a^2} dx$$by defining $z=e^{ix}$ we have$$int_{-pi}^pi {sin xsin nxover 1-2acos x+a^2} dx{=oint_{|z|=1} {{1over 2i}(z-z^{-1}){1over 2i}(z^{n}-z^{-n})over 1-aleft(z+{1over z}right)+a^2} {dzover iz}\=-{1over 4}oint_{|z|=1} {(z-z^{-1})(z^{n}-z^{-n})over 1-aleft(z+{1over z}right)+a^2} {dzover iz}\=-{1over 4}oint_{|z|=1} {(z^2-1)(z^{2n}-1)over z^n((1+a^2)z-aleft(z^2+1right))} {dzover iz}\={1over 4ai}oint_{|z|=1} {(z^2-1)(z^{2n}-1)over z^{n+1}(z-a)left(z-{1over a}right)} {dz}}$$with the singularities of ${(z^2-1)(z^{2n}-1)over z^{n+1}(z-a)left(z-{1over a}right)}$ falling in $z={1over a},0,a$ when $ane 1$. For $r=1$, the only singularity exists in $z=0$.







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          answered 2 days ago









          Mostafa AyazMostafa Ayaz

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              1












              $begingroup$

              $$I=int_{0}^{pi} frac {sin x sin n x}{1-2a cos x+a^2} dx ~~~~(1)$$ Integrate it by parts taking $sin x$ as the first function, we get
              $$I=-frac{n}{2a}int_{0}^{pi} cos n x ln[1-2a cos x +a^2] dx ~~~~(2).$$
              Let $y=cos x+i sin x$, then $$f(x)=ln(1-2acos x+a^2)=ln(1-ay)+ln(1-a/y).$$
              If $a^2<1,$ then $$f(x)=-2[a cos x + frac{a^2cos 2 x}{2}+frac{a^3 cos 3 x}{3}+...]~~~~(3)$$



              Using (3) in (2) and using the property that $$int_{0}^{pi} cos m x cos n x dx=frac{pi}{2}delta_{m,n}$$ we get
              $$I=frac{n}{a} frac{a^n}{n} frac{pi}{2}=frac{pi a^{n-1}}{2},~~~ a^2 <1.$$






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              $endgroup$




















                1












                $begingroup$

                $$I=int_{0}^{pi} frac {sin x sin n x}{1-2a cos x+a^2} dx ~~~~(1)$$ Integrate it by parts taking $sin x$ as the first function, we get
                $$I=-frac{n}{2a}int_{0}^{pi} cos n x ln[1-2a cos x +a^2] dx ~~~~(2).$$
                Let $y=cos x+i sin x$, then $$f(x)=ln(1-2acos x+a^2)=ln(1-ay)+ln(1-a/y).$$
                If $a^2<1,$ then $$f(x)=-2[a cos x + frac{a^2cos 2 x}{2}+frac{a^3 cos 3 x}{3}+...]~~~~(3)$$



                Using (3) in (2) and using the property that $$int_{0}^{pi} cos m x cos n x dx=frac{pi}{2}delta_{m,n}$$ we get
                $$I=frac{n}{a} frac{a^n}{n} frac{pi}{2}=frac{pi a^{n-1}}{2},~~~ a^2 <1.$$






                share|cite|improve this answer











                $endgroup$


















                  1












                  1








                  1





                  $begingroup$

                  $$I=int_{0}^{pi} frac {sin x sin n x}{1-2a cos x+a^2} dx ~~~~(1)$$ Integrate it by parts taking $sin x$ as the first function, we get
                  $$I=-frac{n}{2a}int_{0}^{pi} cos n x ln[1-2a cos x +a^2] dx ~~~~(2).$$
                  Let $y=cos x+i sin x$, then $$f(x)=ln(1-2acos x+a^2)=ln(1-ay)+ln(1-a/y).$$
                  If $a^2<1,$ then $$f(x)=-2[a cos x + frac{a^2cos 2 x}{2}+frac{a^3 cos 3 x}{3}+...]~~~~(3)$$



                  Using (3) in (2) and using the property that $$int_{0}^{pi} cos m x cos n x dx=frac{pi}{2}delta_{m,n}$$ we get
                  $$I=frac{n}{a} frac{a^n}{n} frac{pi}{2}=frac{pi a^{n-1}}{2},~~~ a^2 <1.$$






                  share|cite|improve this answer











                  $endgroup$



                  $$I=int_{0}^{pi} frac {sin x sin n x}{1-2a cos x+a^2} dx ~~~~(1)$$ Integrate it by parts taking $sin x$ as the first function, we get
                  $$I=-frac{n}{2a}int_{0}^{pi} cos n x ln[1-2a cos x +a^2] dx ~~~~(2).$$
                  Let $y=cos x+i sin x$, then $$f(x)=ln(1-2acos x+a^2)=ln(1-ay)+ln(1-a/y).$$
                  If $a^2<1,$ then $$f(x)=-2[a cos x + frac{a^2cos 2 x}{2}+frac{a^3 cos 3 x}{3}+...]~~~~(3)$$



                  Using (3) in (2) and using the property that $$int_{0}^{pi} cos m x cos n x dx=frac{pi}{2}delta_{m,n}$$ we get
                  $$I=frac{n}{a} frac{a^n}{n} frac{pi}{2}=frac{pi a^{n-1}}{2},~~~ a^2 <1.$$







                  share|cite|improve this answer














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                  share|cite|improve this answer








                  edited 2 days ago

























                  answered 2 days ago









                  Dr Zafar Ahmed DScDr Zafar Ahmed DSc

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