$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)},dx$
$begingroup$
I have been trying to solve the integral below using contour integration.
$$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)}dx, quad a>1.$$
I'd appreciate to see how you would solve it, since the fact that it goes to only $pi$ and not to $2pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.
complex-analysis contour-integration
edited Jan 3 at 18:36
Bernard
119k740113
asked Jan 3 at 16:32
SamSam
382114
$endgroup$
add a comment |
$begingroup$
I have been trying to solve the integral below using contour integration.
$$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)}dx, quad a>1.$$
I'd appreciate to see how you would solve it, since the fact that it goes to only $pi$ and not to $2pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.
complex-analysis contour-integration
edited Jan 3 at 18:36
Bernard
119k740113
asked Jan 3 at 16:32
SamSam
382114
$endgroup$
1
$begingroup$
I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
$endgroup$
– Robert Israel
Jan 3 at 16:33
$begingroup$
Oh true, ill give that a try. Thanks
$endgroup$
– Sam
Jan 3 at 16:39
$begingroup$
Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 16:42
add a comment |
$begingroup$
I have been trying to solve the integral below using contour integration.
$$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)}dx, quad a>1.$$
I'd appreciate to see how you would solve it, since the fact that it goes to only $pi$ and not to $2pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.
complex-analysis contour-integration
edited Jan 3 at 18:36
Bernard
119k740113
asked Jan 3 at 16:32
SamSam
382114
$endgroup$
I have been trying to solve the integral below using contour integration.
$$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)}dx, quad a>1.$$
I'd appreciate to see how you would solve it, since the fact that it goes to only $pi$ and not to $2pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.
complex-analysis contour-integration
complex-analysis contour-integration
edited Jan 3 at 18:36
Bernard
119k740113
asked Jan 3 at 16:32
SamSam
382114
edited Jan 3 at 18:36
Bernard
119k740113
asked Jan 3 at 16:32
SamSam
382114
edited Jan 3 at 18:36
Bernard
119k740113
edited Jan 3 at 18:36
Bernard
119k740113
edited Jan 3 at 18:36
Bernard
119k740113
119k740113
asked Jan 3 at 16:32
SamSam
382114
asked Jan 3 at 16:32
SamSam
382114
asked Jan 3 at 16:32
SamSam
382114
382114
1
$begingroup$
I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
$endgroup$
– Robert Israel
Jan 3 at 16:33
$begingroup$
Oh true, ill give that a try. Thanks
$endgroup$
– Sam
Jan 3 at 16:39
$begingroup$
Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 16:42
add a comment |
1
$begingroup$
I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
$endgroup$
– Robert Israel
Jan 3 at 16:33
$begingroup$
Oh true, ill give that a try. Thanks
$endgroup$
– Sam
Jan 3 at 16:39
$begingroup$
Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 16:42
1
1
$begingroup$
I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
$endgroup$
– Robert Israel
Jan 3 at 16:33
$begingroup$
I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
$endgroup$
– Robert Israel
Jan 3 at 16:33
$begingroup$
Oh true, ill give that a try. Thanks
$endgroup$
– Sam
Jan 3 at 16:39
$begingroup$
Oh true, ill give that a try. Thanks
$endgroup$
– Sam
Jan 3 at 16:39
$begingroup$
Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 16:42
$begingroup$
Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 16:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The chances of me saying this were high: let us avoid contour integration.
By symmetry
$$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
equals
$$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
or
$$ frac{pi}{a+sqrt{a^2-1}} $$
by partial fraction decomposition.
answered Jan 3 at 16:51
Jack D'AurizioJack D'Aurizio
1
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
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oldest
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$begingroup$
The chances of me saying this were high: let us avoid contour integration.
By symmetry
$$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
equals
$$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
or
$$ frac{pi}{a+sqrt{a^2-1}} $$
by partial fraction decomposition.
answered Jan 3 at 16:51
Jack D'AurizioJack D'Aurizio
1
$endgroup$
add a comment |
$begingroup$
The chances of me saying this were high: let us avoid contour integration.
By symmetry
$$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
equals
$$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
or
$$ frac{pi}{a+sqrt{a^2-1}} $$
by partial fraction decomposition.
answered Jan 3 at 16:51
Jack D'AurizioJack D'Aurizio
1
$endgroup$
add a comment |
$begingroup$
The chances of me saying this were high: let us avoid contour integration.
By symmetry
$$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
equals
$$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
or
$$ frac{pi}{a+sqrt{a^2-1}} $$
by partial fraction decomposition.
answered Jan 3 at 16:51
Jack D'AurizioJack D'Aurizio
1
$endgroup$
The chances of me saying this were high: let us avoid contour integration.
By symmetry
$$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
equals
$$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
or
$$ frac{pi}{a+sqrt{a^2-1}} $$
by partial fraction decomposition.
answered Jan 3 at 16:51
Jack D'AurizioJack D'Aurizio
1
answered Jan 3 at 16:51
Jack D'AurizioJack D'Aurizio
1
answered Jan 3 at 16:51
Jack D'AurizioJack D'Aurizio
1
answered Jan 3 at 16:51
Jack D'AurizioJack D'Aurizio
1
1
add a comment |
add a comment |
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$begingroup$
I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
$endgroup$
– Robert Israel
Jan 3 at 16:33
$begingroup$
Oh true, ill give that a try. Thanks
$endgroup$
– Sam
Jan 3 at 16:39
$begingroup$
Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 16:42