What is the technique that should be used to evaluate $int_0^2pi (sin (ksin theta))^2 ,dtheta $?

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$$int_0^2pi (sin (ksin theta))^2 ,dtheta $$ Where $k$ is a real constant.




I searched for it and found out that it 'does not have an integral
I tried using a substitution for sinθ but then both limits change to zero



expressible with elementary functions'. What does this mean? If so how to solve it?










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  • I tried using a substitution for $sintheta$ but then both limits change to zero.
    – VJay
    Sep 1 at 8:08











  • Include your work in the question not in comments
    – Deepesh Meena
    Sep 1 at 9:07










  • Your substitution must be bijective on the bounds of the integral for it to be valid
    – aidangallagher4
    Sep 1 at 9:18










  • have you tried to use the identity $(sin t)^2=1/2(1-sin t)$ on the following webpage are some other that could help you to better undertand the problem sosmath.com/trig/prodform/prodform.html
    – Adrián Hinojosa Calleja
    Sep 1 at 9:24










  • en.m.wikipedia.org/wiki/Elementary_function
    – Adrián Hinojosa Calleja
    Sep 1 at 9:28














up vote
1
down vote

favorite













$$int_0^2pi (sin (ksin theta))^2 ,dtheta $$ Where $k$ is a real constant.




I searched for it and found out that it 'does not have an integral
I tried using a substitution for sinθ but then both limits change to zero



expressible with elementary functions'. What does this mean? If so how to solve it?










share|cite|improve this question























  • I tried using a substitution for $sintheta$ but then both limits change to zero.
    – VJay
    Sep 1 at 8:08











  • Include your work in the question not in comments
    – Deepesh Meena
    Sep 1 at 9:07










  • Your substitution must be bijective on the bounds of the integral for it to be valid
    – aidangallagher4
    Sep 1 at 9:18










  • have you tried to use the identity $(sin t)^2=1/2(1-sin t)$ on the following webpage are some other that could help you to better undertand the problem sosmath.com/trig/prodform/prodform.html
    – Adrián Hinojosa Calleja
    Sep 1 at 9:24










  • en.m.wikipedia.org/wiki/Elementary_function
    – Adrián Hinojosa Calleja
    Sep 1 at 9:28












up vote
1
down vote

favorite









up vote
1
down vote

favorite












$$int_0^2pi (sin (ksin theta))^2 ,dtheta $$ Where $k$ is a real constant.




I searched for it and found out that it 'does not have an integral
I tried using a substitution for sinθ but then both limits change to zero



expressible with elementary functions'. What does this mean? If so how to solve it?










share|cite|improve this question
















$$int_0^2pi (sin (ksin theta))^2 ,dtheta $$ Where $k$ is a real constant.




I searched for it and found out that it 'does not have an integral
I tried using a substitution for sinθ but then both limits change to zero



expressible with elementary functions'. What does this mean? If so how to solve it?







integration trigonometry






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share|cite|improve this question













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edited Sep 1 at 10:06









Blue

44.1k868141




44.1k868141










asked Sep 1 at 8:00









VJay

113




113











  • I tried using a substitution for $sintheta$ but then both limits change to zero.
    – VJay
    Sep 1 at 8:08











  • Include your work in the question not in comments
    – Deepesh Meena
    Sep 1 at 9:07










  • Your substitution must be bijective on the bounds of the integral for it to be valid
    – aidangallagher4
    Sep 1 at 9:18










  • have you tried to use the identity $(sin t)^2=1/2(1-sin t)$ on the following webpage are some other that could help you to better undertand the problem sosmath.com/trig/prodform/prodform.html
    – Adrián Hinojosa Calleja
    Sep 1 at 9:24










  • en.m.wikipedia.org/wiki/Elementary_function
    – Adrián Hinojosa Calleja
    Sep 1 at 9:28
















  • I tried using a substitution for $sintheta$ but then both limits change to zero.
    – VJay
    Sep 1 at 8:08











  • Include your work in the question not in comments
    – Deepesh Meena
    Sep 1 at 9:07










  • Your substitution must be bijective on the bounds of the integral for it to be valid
    – aidangallagher4
    Sep 1 at 9:18










  • have you tried to use the identity $(sin t)^2=1/2(1-sin t)$ on the following webpage are some other that could help you to better undertand the problem sosmath.com/trig/prodform/prodform.html
    – Adrián Hinojosa Calleja
    Sep 1 at 9:24










  • en.m.wikipedia.org/wiki/Elementary_function
    – Adrián Hinojosa Calleja
    Sep 1 at 9:28















I tried using a substitution for $sintheta$ but then both limits change to zero.
– VJay
Sep 1 at 8:08





I tried using a substitution for $sintheta$ but then both limits change to zero.
– VJay
Sep 1 at 8:08













Include your work in the question not in comments
– Deepesh Meena
Sep 1 at 9:07




Include your work in the question not in comments
– Deepesh Meena
Sep 1 at 9:07












Your substitution must be bijective on the bounds of the integral for it to be valid
– aidangallagher4
Sep 1 at 9:18




Your substitution must be bijective on the bounds of the integral for it to be valid
– aidangallagher4
Sep 1 at 9:18












have you tried to use the identity $(sin t)^2=1/2(1-sin t)$ on the following webpage are some other that could help you to better undertand the problem sosmath.com/trig/prodform/prodform.html
– Adrián Hinojosa Calleja
Sep 1 at 9:24




have you tried to use the identity $(sin t)^2=1/2(1-sin t)$ on the following webpage are some other that could help you to better undertand the problem sosmath.com/trig/prodform/prodform.html
– Adrián Hinojosa Calleja
Sep 1 at 9:24












en.m.wikipedia.org/wiki/Elementary_function
– Adrián Hinojosa Calleja
Sep 1 at 9:28




en.m.wikipedia.org/wiki/Elementary_function
– Adrián Hinojosa Calleja
Sep 1 at 9:28










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Hint:
$$int_0^2pi (sin (ksin theta))^2 ,dtheta=int_0^2pi dfrac12left(1-cos (2ksin theta)right) ,dtheta$$
where
$$int_0^pi cos(xsin t) dt=pi J_0(x)$$
and $J_0$ is Bessel function of order $0$.






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    1 Answer
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    up vote
    3
    down vote













    Hint:
    $$int_0^2pi (sin (ksin theta))^2 ,dtheta=int_0^2pi dfrac12left(1-cos (2ksin theta)right) ,dtheta$$
    where
    $$int_0^pi cos(xsin t) dt=pi J_0(x)$$
    and $J_0$ is Bessel function of order $0$.






    share|cite|improve this answer
























      up vote
      3
      down vote













      Hint:
      $$int_0^2pi (sin (ksin theta))^2 ,dtheta=int_0^2pi dfrac12left(1-cos (2ksin theta)right) ,dtheta$$
      where
      $$int_0^pi cos(xsin t) dt=pi J_0(x)$$
      and $J_0$ is Bessel function of order $0$.






      share|cite|improve this answer






















        up vote
        3
        down vote










        up vote
        3
        down vote









        Hint:
        $$int_0^2pi (sin (ksin theta))^2 ,dtheta=int_0^2pi dfrac12left(1-cos (2ksin theta)right) ,dtheta$$
        where
        $$int_0^pi cos(xsin t) dt=pi J_0(x)$$
        and $J_0$ is Bessel function of order $0$.






        share|cite|improve this answer












        Hint:
        $$int_0^2pi (sin (ksin theta))^2 ,dtheta=int_0^2pi dfrac12left(1-cos (2ksin theta)right) ,dtheta$$
        where
        $$int_0^pi cos(xsin t) dt=pi J_0(x)$$
        and $J_0$ is Bessel function of order $0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 1 at 9:58









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