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 sinø but then both limits change to zero
expressible with elementary functions'. What does this mean? If so how to solve it?
integration trigonometry
 |Â
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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 sinø but then both limits change to zero
expressible with elementary functions'. What does this mean? If so how to solve it?
integration trigonometry
I tried using a substitution for $sintheta$ but then both limits change to zero.
â VJay
Sep 1 at 8:08
Include your work in the question not in comments
â Deepesh Meena
Sep 1 at 9:07
Your substitution must be bijective on the bounds of the integral for it to be valid
â 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
â Adrián Hinojosa Calleja
Sep 1 at 9:24
en.m.wikipedia.org/wiki/Elementary_function
â Adrián Hinojosa Calleja
Sep 1 at 9:28
 |Â
show 1 more comment
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 sinø but then both limits change to zero
expressible with elementary functions'. What does this mean? If so how to solve it?
integration trigonometry
$$int_0^2pi (sin (ksin theta))^2 ,dtheta $$ 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 sinø but then both limits change to zero
expressible with elementary functions'. What does this mean? If so how to solve it?
integration trigonometry
integration trigonometry
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.
â VJay
Sep 1 at 8:08
Include your work in the question not in comments
â Deepesh Meena
Sep 1 at 9:07
Your substitution must be bijective on the bounds of the integral for it to be valid
â 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
â Adrián Hinojosa Calleja
Sep 1 at 9:24
en.m.wikipedia.org/wiki/Elementary_function
â Adrián Hinojosa Calleja
Sep 1 at 9:28
 |Â
show 1 more comment
I tried using a substitution for $sintheta$ but then both limits change to zero.
â VJay
Sep 1 at 8:08
Include your work in the question not in comments
â Deepesh Meena
Sep 1 at 9:07
Your substitution must be bijective on the bounds of the integral for it to be valid
â 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
â Adrián Hinojosa Calleja
Sep 1 at 9:24
en.m.wikipedia.org/wiki/Elementary_function
â Adrián Hinojosa Calleja
Sep 1 at 9:28
I tried using a substitution for $sintheta$ but then both limits change to zero.
â VJay
Sep 1 at 8:08
I tried using a substitution for $sintheta$ but then both limits change to zero.
â VJay
Sep 1 at 8:08
Include your work in the question not in comments
â Deepesh Meena
Sep 1 at 9:07
Include your work in the question not in comments
â Deepesh Meena
Sep 1 at 9:07
Your substitution must be bijective on the bounds of the integral for it to be valid
â aidangallagher4
Sep 1 at 9:18
Your substitution must be bijective on the bounds of the integral for it to be valid
â 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
â 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
â Adrián Hinojosa Calleja
Sep 1 at 9:24
en.m.wikipedia.org/wiki/Elementary_function
â Adrián Hinojosa Calleja
Sep 1 at 9:28
en.m.wikipedia.org/wiki/Elementary_function
â Adrián Hinojosa Calleja
Sep 1 at 9:28
 |Â
show 1 more comment
1 Answer
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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 $0$.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $0$.
add a comment |Â
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 $0$.
add a comment |Â
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 $0$.
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 $0$.
answered Sep 1 at 9:58
Nosrati
22.1k61747
22.1k61747
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I tried using a substitution for $sintheta$ but then both limits change to zero.
â VJay
Sep 1 at 8:08
Include your work in the question not in comments
â Deepesh Meena
Sep 1 at 9:07
Your substitution must be bijective on the bounds of the integral for it to be valid
â 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
â Adrián Hinojosa Calleja
Sep 1 at 9:24
en.m.wikipedia.org/wiki/Elementary_function
â Adrián Hinojosa Calleja
Sep 1 at 9:28