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How to evaluate $int_{0}^{2pi}frac{1}{1+sin^{2}(theta)}dtheta$
I'm pretty sure the idea is to interpret this integral as a contour integral over a closed path in the complex plane.
$$I = int_{0}^{2pi}frac{1}{1+sin^{2}(theta)}dtheta$$
We take $C = {z in mathbb{C} : |z| = 1}$, which is positively oriented.
$$z = e^{itheta}$$
$$frac{dz}{dtheta} = ie^{itheta} = iz rightarrow dtheta = frac{dz}{iz} $$
$$sin(theta) = frac{e^{itheta}-e^{-itheta}}{2i} = frac{z-z^{-1}}{2i} $$
$$sin^2(theta) = frac{z^2 - 2zz^{-1} + z^{-2}}{(2i)^2} = frac{z^2 - 2 + z^{-2}}{-4} $$
After some manipulation:
$$I = frac{4}{i}oint_{C}frac{z}{-z^4+6z^2-1}dz$$
After that one should obtain the roots of the polynomial in the denominator and solve the integral either by residues or using Cauchy's formula. However, I'm not sure how to find the roots. This was in the context of an exam, so I assume something must be wrong if the roots are too difficult to find, or if they are not "nice" numbers.
Is there a more clever way of going about this? I tried calculating the residue at infinity, but $frac{1}{z^{2}}f(frac{1}{z})$ yields the exact same function and it doesn't get any easier.
Thanks.
integration complex-analysis definite-integrals
edited Nov 21 '18 at 0:26
Bernard
118k639112
asked Nov 21 '18 at 0:14
Juanma Eloy
5511516
add a comment |
I'm pretty sure the idea is to interpret this integral as a contour integral over a closed path in the complex plane.
$$I = int_{0}^{2pi}frac{1}{1+sin^{2}(theta)}dtheta$$
We take $C = {z in mathbb{C} : |z| = 1}$, which is positively oriented.
$$z = e^{itheta}$$
$$frac{dz}{dtheta} = ie^{itheta} = iz rightarrow dtheta = frac{dz}{iz} $$
$$sin(theta) = frac{e^{itheta}-e^{-itheta}}{2i} = frac{z-z^{-1}}{2i} $$
$$sin^2(theta) = frac{z^2 - 2zz^{-1} + z^{-2}}{(2i)^2} = frac{z^2 - 2 + z^{-2}}{-4} $$
After some manipulation:
$$I = frac{4}{i}oint_{C}frac{z}{-z^4+6z^2-1}dz$$
After that one should obtain the roots of the polynomial in the denominator and solve the integral either by residues or using Cauchy's formula. However, I'm not sure how to find the roots. This was in the context of an exam, so I assume something must be wrong if the roots are too difficult to find, or if they are not "nice" numbers.
Is there a more clever way of going about this? I tried calculating the residue at infinity, but $frac{1}{z^{2}}f(frac{1}{z})$ yields the exact same function and it doesn't get any easier.
Thanks.
integration complex-analysis definite-integrals
edited Nov 21 '18 at 0:26
Bernard
118k639112
asked Nov 21 '18 at 0:14
Juanma Eloy
5511516
1
$z^4 = (z^2)^2$ so you can use the quadratic formula to get the roots.
– Winther
Nov 21 '18 at 0:19
1
Note that $z^4 - 6z^2 + 1 = (z^2)^2 - 6(z^2) + 1$ can be factored using the quadratic formula. Alternatively, $u = z^2$ works as a $u$-substitution.
– Omnomnomnom
Nov 21 '18 at 0:20
1
You could write $I = int_0^{2pi} frac{1}{1 + frac{1}{2} (1 - cos(2theta))} dtheta = int_0^{2pi} frac{2}{3 - cos(2theta)}dtheta = int_0^{4pi} frac{1}{3 - cos alpha}dalpha = 2 int_0^{2pi} frac{1}{3-cos alpha}dalpha$ and then evaluate that by contour integration.
– Daniel Schepler
Nov 21 '18 at 1:58
add a comment |
I'm pretty sure the idea is to interpret this integral as a contour integral over a closed path in the complex plane.
$$I = int_{0}^{2pi}frac{1}{1+sin^{2}(theta)}dtheta$$
We take $C = {z in mathbb{C} : |z| = 1}$, which is positively oriented.
$$z = e^{itheta}$$
$$frac{dz}{dtheta} = ie^{itheta} = iz rightarrow dtheta = frac{dz}{iz} $$
$$sin(theta) = frac{e^{itheta}-e^{-itheta}}{2i} = frac{z-z^{-1}}{2i} $$
$$sin^2(theta) = frac{z^2 - 2zz^{-1} + z^{-2}}{(2i)^2} = frac{z^2 - 2 + z^{-2}}{-4} $$
After some manipulation:
$$I = frac{4}{i}oint_{C}frac{z}{-z^4+6z^2-1}dz$$
After that one should obtain the roots of the polynomial in the denominator and solve the integral either by residues or using Cauchy's formula. However, I'm not sure how to find the roots. This was in the context of an exam, so I assume something must be wrong if the roots are too difficult to find, or if they are not "nice" numbers.
Is there a more clever way of going about this? I tried calculating the residue at infinity, but $frac{1}{z^{2}}f(frac{1}{z})$ yields the exact same function and it doesn't get any easier.
Thanks.
integration complex-analysis definite-integrals
edited Nov 21 '18 at 0:26
Bernard
118k639112
asked Nov 21 '18 at 0:14
Juanma Eloy
5511516
I'm pretty sure the idea is to interpret this integral as a contour integral over a closed path in the complex plane.
$$I = int_{0}^{2pi}frac{1}{1+sin^{2}(theta)}dtheta$$
We take $C = {z in mathbb{C} : |z| = 1}$, which is positively oriented.
$$z = e^{itheta}$$
$$frac{dz}{dtheta} = ie^{itheta} = iz rightarrow dtheta = frac{dz}{iz} $$
$$sin(theta) = frac{e^{itheta}-e^{-itheta}}{2i} = frac{z-z^{-1}}{2i} $$
$$sin^2(theta) = frac{z^2 - 2zz^{-1} + z^{-2}}{(2i)^2} = frac{z^2 - 2 + z^{-2}}{-4} $$
After some manipulation:
$$I = frac{4}{i}oint_{C}frac{z}{-z^4+6z^2-1}dz$$
After that one should obtain the roots of the polynomial in the denominator and solve the integral either by residues or using Cauchy's formula. However, I'm not sure how to find the roots. This was in the context of an exam, so I assume something must be wrong if the roots are too difficult to find, or if they are not "nice" numbers.
Is there a more clever way of going about this? I tried calculating the residue at infinity, but $frac{1}{z^{2}}f(frac{1}{z})$ yields the exact same function and it doesn't get any easier.
Thanks.
integration complex-analysis definite-integrals
integration complex-analysis definite-integrals
edited Nov 21 '18 at 0:26
Bernard
118k639112
asked Nov 21 '18 at 0:14
Juanma Eloy
5511516
edited Nov 21 '18 at 0:26
Bernard
118k639112
asked Nov 21 '18 at 0:14
Juanma Eloy
5511516
edited Nov 21 '18 at 0:26
Bernard
118k639112
edited Nov 21 '18 at 0:26
Bernard
118k639112
edited Nov 21 '18 at 0:26
Bernard
118k639112
118k639112
asked Nov 21 '18 at 0:14
Juanma Eloy
5511516
asked Nov 21 '18 at 0:14
Juanma Eloy
5511516
asked Nov 21 '18 at 0:14
Juanma Eloy
5511516
5511516
1
$z^4 = (z^2)^2$ so you can use the quadratic formula to get the roots.
– Winther
Nov 21 '18 at 0:19
1
Note that $z^4 - 6z^2 + 1 = (z^2)^2 - 6(z^2) + 1$ can be factored using the quadratic formula. Alternatively, $u = z^2$ works as a $u$-substitution.
– Omnomnomnom
Nov 21 '18 at 0:20
1
You could write $I = int_0^{2pi} frac{1}{1 + frac{1}{2} (1 - cos(2theta))} dtheta = int_0^{2pi} frac{2}{3 - cos(2theta)}dtheta = int_0^{4pi} frac{1}{3 - cos alpha}dalpha = 2 int_0^{2pi} frac{1}{3-cos alpha}dalpha$ and then evaluate that by contour integration.
– Daniel Schepler
Nov 21 '18 at 1:58
add a comment |
1
$z^4 = (z^2)^2$ so you can use the quadratic formula to get the roots.
– Winther
Nov 21 '18 at 0:19
1
Note that $z^4 - 6z^2 + 1 = (z^2)^2 - 6(z^2) + 1$ can be factored using the quadratic formula. Alternatively, $u = z^2$ works as a $u$-substitution.
– Omnomnomnom
Nov 21 '18 at 0:20
1
You could write $I = int_0^{2pi} frac{1}{1 + frac{1}{2} (1 - cos(2theta))} dtheta = int_0^{2pi} frac{2}{3 - cos(2theta)}dtheta = int_0^{4pi} frac{1}{3 - cos alpha}dalpha = 2 int_0^{2pi} frac{1}{3-cos alpha}dalpha$ and then evaluate that by contour integration.
– Daniel Schepler
Nov 21 '18 at 1:58
1
1
$z^4 = (z^2)^2$ so you can use the quadratic formula to get the roots.
– Winther
Nov 21 '18 at 0:19
$z^4 = (z^2)^2$ so you can use the quadratic formula to get the roots.
– Winther
Nov 21 '18 at 0:19
1
1
Note that $z^4 - 6z^2 + 1 = (z^2)^2 - 6(z^2) + 1$ can be factored using the quadratic formula. Alternatively, $u = z^2$ works as a $u$-substitution.
– Omnomnomnom
Nov 21 '18 at 0:20
Note that $z^4 - 6z^2 + 1 = (z^2)^2 - 6(z^2) + 1$ can be factored using the quadratic formula. Alternatively, $u = z^2$ works as a $u$-substitution.
– Omnomnomnom
Nov 21 '18 at 0:20
1
1
You could write $I = int_0^{2pi} frac{1}{1 + frac{1}{2} (1 - cos(2theta))} dtheta = int_0^{2pi} frac{2}{3 - cos(2theta)}dtheta = int_0^{4pi} frac{1}{3 - cos alpha}dalpha = 2 int_0^{2pi} frac{1}{3-cos alpha}dalpha$ and then evaluate that by contour integration.
– Daniel Schepler
Nov 21 '18 at 1:58
You could write $I = int_0^{2pi} frac{1}{1 + frac{1}{2} (1 - cos(2theta))} dtheta = int_0^{2pi} frac{2}{3 - cos(2theta)}dtheta = int_0^{4pi} frac{1}{3 - cos alpha}dalpha = 2 int_0^{2pi} frac{1}{3-cos alpha}dalpha$ and then evaluate that by contour integration.
– Daniel Schepler
Nov 21 '18 at 1:58
add a comment |
3 Answers
3
active
oldest
votes
Contour integration is a chance, but not the only one. For instance, symmetry gives
$$ int_{0}^{2pi}frac{dtheta}{1+sin^2theta}=4int_{0}^{pi/2}frac{dtheta}{1+sin^2theta}=4int_{0}^{pi/2}frac{dtheta}{1+cos^2theta}stackrel{thetamapstoarctan u}{=}4int_{0}^{+infty}frac{du}{2+u^2} $$
and the last integral clearly equals $color{red}{pisqrt{2}}$.
answered Nov 21 '18 at 0:35
Jack D'Aurizio
287k33280657
“clearly” not so clear
– Lucas Henrique
Nov 21 '18 at 8:29
@LucasHenrique: well, $int_{a}^{b}frac{dx}{1+x^2}=arctan(b)-arctan(a)$ and by letting $u=sqrt{2},x$...
– Jack D'Aurizio
Nov 21 '18 at 16:42
add a comment |
$$I=intfrac{mathrm{d}x}{1+sin^2x}$$
Recall that $sin x=frac{tan x}{sec x}$:
$$I=intfrac{mathrm{d}x}{1+frac{tan^2x}{sec^2x}}$$
$$I=intfrac{sec^2x mathrm{d}x}{sec^2x+tan^2x}$$
$$I=intfrac{sec^2x mathrm{d}x}{1+2tan^2x}$$
$u=tan x$:
$$I=intfrac{mathrm{d}u}{1+2u^2}$$
$u=frac1{sqrt{2}}tan w$:
$$I=frac1{sqrt{2}}intfrac{sec^2wmathrm{d}w}{1+tan^2w}$$
$$I=frac1{sqrt{2}}w$$
$$I=frac1{sqrt{2}}arctansqrt{2}tan x$$
$$Ibigg|_0^{2pi}=pisqrt2$$
answered Nov 21 '18 at 1:49
clathratus
3,243331
add a comment |
How about using Weierstrauss substitution.
$$I=int_0^{2pi}frac{1}{1+sin^2(x)}dx$$
$t=tanfrac{x}{2}$ so it becomes:
$$I=4int_0^inftyfrac{1}{1+left(frac{2t}{1+t^2}right)^2}frac{dt}{1+t^2}=4int_0^inftyfrac{1+t^2}{(1+t^2)^2+(2t)^2}dt$$
$$=4int_0^inftyfrac{t^2+1}{t^4+6t^2+1}dt=4int_0^inftyfrac{t^2+1}{(t^2+3-2sqrt{2})(t^2+3+2sqrt{2})}dt$$
now using partial fractions:
$$frac{t^2+1}{(t^2+3-2sqrt{2})(t^2+3+2sqrt{2})}=frac{At+B}{t^2+3-2sqrt{2}}+frac{Ct+D}{t^2+3+2sqrt{2}}$$
giving us the simultaneous equations:
$$A+C=0$$$$B+D=1$$$$(3+2sqrt{2})A+(3-2sqrt{2})C=0$$$$(3+2sqrt{2})B+(3-2sqrt{2})D=1$$
Both $A$ and $C$ come out as $0$ so some form of $tan$ substitution can be used to solve the two integrals that are formed.
answered Nov 21 '18 at 3:34
Henry Lee
1,773218
1
You can simplify the algebra by first using the cosine double angle formula $sin^2(x) = (1/2)(1 - cos(2x))$ and then a change of formula $u = 2x$ and then apply the Weierstrass substitution.
– DavidG
Nov 21 '18 at 4:20
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Contour integration is a chance, but not the only one. For instance, symmetry gives
$$ int_{0}^{2pi}frac{dtheta}{1+sin^2theta}=4int_{0}^{pi/2}frac{dtheta}{1+sin^2theta}=4int_{0}^{pi/2}frac{dtheta}{1+cos^2theta}stackrel{thetamapstoarctan u}{=}4int_{0}^{+infty}frac{du}{2+u^2} $$
and the last integral clearly equals $color{red}{pisqrt{2}}$.
answered Nov 21 '18 at 0:35
Jack D'Aurizio
287k33280657
“clearly” not so clear
– Lucas Henrique
Nov 21 '18 at 8:29
@LucasHenrique: well, $int_{a}^{b}frac{dx}{1+x^2}=arctan(b)-arctan(a)$ and by letting $u=sqrt{2},x$...
– Jack D'Aurizio
Nov 21 '18 at 16:42
add a comment |
Contour integration is a chance, but not the only one. For instance, symmetry gives
$$ int_{0}^{2pi}frac{dtheta}{1+sin^2theta}=4int_{0}^{pi/2}frac{dtheta}{1+sin^2theta}=4int_{0}^{pi/2}frac{dtheta}{1+cos^2theta}stackrel{thetamapstoarctan u}{=}4int_{0}^{+infty}frac{du}{2+u^2} $$
and the last integral clearly equals $color{red}{pisqrt{2}}$.
answered Nov 21 '18 at 0:35
Jack D'Aurizio
287k33280657
“clearly” not so clear
– Lucas Henrique
Nov 21 '18 at 8:29
@LucasHenrique: well, $int_{a}^{b}frac{dx}{1+x^2}=arctan(b)-arctan(a)$ and by letting $u=sqrt{2},x$...
– Jack D'Aurizio
Nov 21 '18 at 16:42
add a comment |
Contour integration is a chance, but not the only one. For instance, symmetry gives
$$ int_{0}^{2pi}frac{dtheta}{1+sin^2theta}=4int_{0}^{pi/2}frac{dtheta}{1+sin^2theta}=4int_{0}^{pi/2}frac{dtheta}{1+cos^2theta}stackrel{thetamapstoarctan u}{=}4int_{0}^{+infty}frac{du}{2+u^2} $$
and the last integral clearly equals $color{red}{pisqrt{2}}$.
answered Nov 21 '18 at 0:35
Jack D'Aurizio
287k33280657
Contour integration is a chance, but not the only one. For instance, symmetry gives
$$ int_{0}^{2pi}frac{dtheta}{1+sin^2theta}=4int_{0}^{pi/2}frac{dtheta}{1+sin^2theta}=4int_{0}^{pi/2}frac{dtheta}{1+cos^2theta}stackrel{thetamapstoarctan u}{=}4int_{0}^{+infty}frac{du}{2+u^2} $$
and the last integral clearly equals $color{red}{pisqrt{2}}$.
answered Nov 21 '18 at 0:35
Jack D'Aurizio
287k33280657
answered Nov 21 '18 at 0:35
Jack D'Aurizio
287k33280657
answered Nov 21 '18 at 0:35
Jack D'Aurizio
287k33280657
answered Nov 21 '18 at 0:35
Jack D'Aurizio
287k33280657
287k33280657
“clearly” not so clear
– Lucas Henrique
Nov 21 '18 at 8:29
@LucasHenrique: well, $int_{a}^{b}frac{dx}{1+x^2}=arctan(b)-arctan(a)$ and by letting $u=sqrt{2},x$...
– Jack D'Aurizio
Nov 21 '18 at 16:42
add a comment |
“clearly” not so clear
– Lucas Henrique
Nov 21 '18 at 8:29
@LucasHenrique: well, $int_{a}^{b}frac{dx}{1+x^2}=arctan(b)-arctan(a)$ and by letting $u=sqrt{2},x$...
– Jack D'Aurizio
Nov 21 '18 at 16:42
“clearly” not so clear
– Lucas Henrique
Nov 21 '18 at 8:29
“clearly” not so clear
– Lucas Henrique
Nov 21 '18 at 8:29
@LucasHenrique: well, $int_{a}^{b}frac{dx}{1+x^2}=arctan(b)-arctan(a)$ and by letting $u=sqrt{2},x$...
– Jack D'Aurizio
Nov 21 '18 at 16:42
@LucasHenrique: well, $int_{a}^{b}frac{dx}{1+x^2}=arctan(b)-arctan(a)$ and by letting $u=sqrt{2},x$...
– Jack D'Aurizio
Nov 21 '18 at 16:42
add a comment |
$$I=intfrac{mathrm{d}x}{1+sin^2x}$$
Recall that $sin x=frac{tan x}{sec x}$:
$$I=intfrac{mathrm{d}x}{1+frac{tan^2x}{sec^2x}}$$
$$I=intfrac{sec^2x mathrm{d}x}{sec^2x+tan^2x}$$
$$I=intfrac{sec^2x mathrm{d}x}{1+2tan^2x}$$
$u=tan x$:
$$I=intfrac{mathrm{d}u}{1+2u^2}$$
$u=frac1{sqrt{2}}tan w$:
$$I=frac1{sqrt{2}}intfrac{sec^2wmathrm{d}w}{1+tan^2w}$$
$$I=frac1{sqrt{2}}w$$
$$I=frac1{sqrt{2}}arctansqrt{2}tan x$$
$$Ibigg|_0^{2pi}=pisqrt2$$
answered Nov 21 '18 at 1:49
clathratus
3,243331
add a comment |
$$I=intfrac{mathrm{d}x}{1+sin^2x}$$
Recall that $sin x=frac{tan x}{sec x}$:
$$I=intfrac{mathrm{d}x}{1+frac{tan^2x}{sec^2x}}$$
$$I=intfrac{sec^2x mathrm{d}x}{sec^2x+tan^2x}$$
$$I=intfrac{sec^2x mathrm{d}x}{1+2tan^2x}$$
$u=tan x$:
$$I=intfrac{mathrm{d}u}{1+2u^2}$$
$u=frac1{sqrt{2}}tan w$:
$$I=frac1{sqrt{2}}intfrac{sec^2wmathrm{d}w}{1+tan^2w}$$
$$I=frac1{sqrt{2}}w$$
$$I=frac1{sqrt{2}}arctansqrt{2}tan x$$
$$Ibigg|_0^{2pi}=pisqrt2$$
answered Nov 21 '18 at 1:49
clathratus
3,243331
add a comment |
$$I=intfrac{mathrm{d}x}{1+sin^2x}$$
Recall that $sin x=frac{tan x}{sec x}$:
$$I=intfrac{mathrm{d}x}{1+frac{tan^2x}{sec^2x}}$$
$$I=intfrac{sec^2x mathrm{d}x}{sec^2x+tan^2x}$$
$$I=intfrac{sec^2x mathrm{d}x}{1+2tan^2x}$$
$u=tan x$:
$$I=intfrac{mathrm{d}u}{1+2u^2}$$
$u=frac1{sqrt{2}}tan w$:
$$I=frac1{sqrt{2}}intfrac{sec^2wmathrm{d}w}{1+tan^2w}$$
$$I=frac1{sqrt{2}}w$$
$$I=frac1{sqrt{2}}arctansqrt{2}tan x$$
$$Ibigg|_0^{2pi}=pisqrt2$$
answered Nov 21 '18 at 1:49
clathratus
3,243331
$$I=intfrac{mathrm{d}x}{1+sin^2x}$$
Recall that $sin x=frac{tan x}{sec x}$:
$$I=intfrac{mathrm{d}x}{1+frac{tan^2x}{sec^2x}}$$
$$I=intfrac{sec^2x mathrm{d}x}{sec^2x+tan^2x}$$
$$I=intfrac{sec^2x mathrm{d}x}{1+2tan^2x}$$
$u=tan x$:
$$I=intfrac{mathrm{d}u}{1+2u^2}$$
$u=frac1{sqrt{2}}tan w$:
$$I=frac1{sqrt{2}}intfrac{sec^2wmathrm{d}w}{1+tan^2w}$$
$$I=frac1{sqrt{2}}w$$
$$I=frac1{sqrt{2}}arctansqrt{2}tan x$$
$$Ibigg|_0^{2pi}=pisqrt2$$
answered Nov 21 '18 at 1:49
clathratus
3,243331
answered Nov 21 '18 at 1:49
clathratus
3,243331
answered Nov 21 '18 at 1:49
clathratus
3,243331
answered Nov 21 '18 at 1:49
clathratus
3,243331
3,243331
add a comment |
add a comment |
How about using Weierstrauss substitution.
$$I=int_0^{2pi}frac{1}{1+sin^2(x)}dx$$
$t=tanfrac{x}{2}$ so it becomes:
$$I=4int_0^inftyfrac{1}{1+left(frac{2t}{1+t^2}right)^2}frac{dt}{1+t^2}=4int_0^inftyfrac{1+t^2}{(1+t^2)^2+(2t)^2}dt$$
$$=4int_0^inftyfrac{t^2+1}{t^4+6t^2+1}dt=4int_0^inftyfrac{t^2+1}{(t^2+3-2sqrt{2})(t^2+3+2sqrt{2})}dt$$
now using partial fractions:
$$frac{t^2+1}{(t^2+3-2sqrt{2})(t^2+3+2sqrt{2})}=frac{At+B}{t^2+3-2sqrt{2}}+frac{Ct+D}{t^2+3+2sqrt{2}}$$
giving us the simultaneous equations:
$$A+C=0$$$$B+D=1$$$$(3+2sqrt{2})A+(3-2sqrt{2})C=0$$$$(3+2sqrt{2})B+(3-2sqrt{2})D=1$$
Both $A$ and $C$ come out as $0$ so some form of $tan$ substitution can be used to solve the two integrals that are formed.
answered Nov 21 '18 at 3:34
Henry Lee
1,773218
1
You can simplify the algebra by first using the cosine double angle formula $sin^2(x) = (1/2)(1 - cos(2x))$ and then a change of formula $u = 2x$ and then apply the Weierstrass substitution.
– DavidG
Nov 21 '18 at 4:20
add a comment |
How about using Weierstrauss substitution.
$$I=int_0^{2pi}frac{1}{1+sin^2(x)}dx$$
$t=tanfrac{x}{2}$ so it becomes:
$$I=4int_0^inftyfrac{1}{1+left(frac{2t}{1+t^2}right)^2}frac{dt}{1+t^2}=4int_0^inftyfrac{1+t^2}{(1+t^2)^2+(2t)^2}dt$$
$$=4int_0^inftyfrac{t^2+1}{t^4+6t^2+1}dt=4int_0^inftyfrac{t^2+1}{(t^2+3-2sqrt{2})(t^2+3+2sqrt{2})}dt$$
now using partial fractions:
$$frac{t^2+1}{(t^2+3-2sqrt{2})(t^2+3+2sqrt{2})}=frac{At+B}{t^2+3-2sqrt{2}}+frac{Ct+D}{t^2+3+2sqrt{2}}$$
giving us the simultaneous equations:
$$A+C=0$$$$B+D=1$$$$(3+2sqrt{2})A+(3-2sqrt{2})C=0$$$$(3+2sqrt{2})B+(3-2sqrt{2})D=1$$
Both $A$ and $C$ come out as $0$ so some form of $tan$ substitution can be used to solve the two integrals that are formed.
answered Nov 21 '18 at 3:34
Henry Lee
1,773218
1
You can simplify the algebra by first using the cosine double angle formula $sin^2(x) = (1/2)(1 - cos(2x))$ and then a change of formula $u = 2x$ and then apply the Weierstrass substitution.
– DavidG
Nov 21 '18 at 4:20
add a comment |
How about using Weierstrauss substitution.
$$I=int_0^{2pi}frac{1}{1+sin^2(x)}dx$$
$t=tanfrac{x}{2}$ so it becomes:
$$I=4int_0^inftyfrac{1}{1+left(frac{2t}{1+t^2}right)^2}frac{dt}{1+t^2}=4int_0^inftyfrac{1+t^2}{(1+t^2)^2+(2t)^2}dt$$
$$=4int_0^inftyfrac{t^2+1}{t^4+6t^2+1}dt=4int_0^inftyfrac{t^2+1}{(t^2+3-2sqrt{2})(t^2+3+2sqrt{2})}dt$$
now using partial fractions:
$$frac{t^2+1}{(t^2+3-2sqrt{2})(t^2+3+2sqrt{2})}=frac{At+B}{t^2+3-2sqrt{2}}+frac{Ct+D}{t^2+3+2sqrt{2}}$$
giving us the simultaneous equations:
$$A+C=0$$$$B+D=1$$$$(3+2sqrt{2})A+(3-2sqrt{2})C=0$$$$(3+2sqrt{2})B+(3-2sqrt{2})D=1$$
Both $A$ and $C$ come out as $0$ so some form of $tan$ substitution can be used to solve the two integrals that are formed.
answered Nov 21 '18 at 3:34
Henry Lee
1,773218
How about using Weierstrauss substitution.
$$I=int_0^{2pi}frac{1}{1+sin^2(x)}dx$$
$t=tanfrac{x}{2}$ so it becomes:
$$I=4int_0^inftyfrac{1}{1+left(frac{2t}{1+t^2}right)^2}frac{dt}{1+t^2}=4int_0^inftyfrac{1+t^2}{(1+t^2)^2+(2t)^2}dt$$
$$=4int_0^inftyfrac{t^2+1}{t^4+6t^2+1}dt=4int_0^inftyfrac{t^2+1}{(t^2+3-2sqrt{2})(t^2+3+2sqrt{2})}dt$$
now using partial fractions:
$$frac{t^2+1}{(t^2+3-2sqrt{2})(t^2+3+2sqrt{2})}=frac{At+B}{t^2+3-2sqrt{2}}+frac{Ct+D}{t^2+3+2sqrt{2}}$$
giving us the simultaneous equations:
$$A+C=0$$$$B+D=1$$$$(3+2sqrt{2})A+(3-2sqrt{2})C=0$$$$(3+2sqrt{2})B+(3-2sqrt{2})D=1$$
Both $A$ and $C$ come out as $0$ so some form of $tan$ substitution can be used to solve the two integrals that are formed.
answered Nov 21 '18 at 3:34
Henry Lee
1,773218
answered Nov 21 '18 at 3:34
Henry Lee
1,773218
answered Nov 21 '18 at 3:34
Henry Lee
1,773218
answered Nov 21 '18 at 3:34
Henry Lee
1,773218
1,773218
1
You can simplify the algebra by first using the cosine double angle formula $sin^2(x) = (1/2)(1 - cos(2x))$ and then a change of formula $u = 2x$ and then apply the Weierstrass substitution.
– DavidG
Nov 21 '18 at 4:20
add a comment |
1
You can simplify the algebra by first using the cosine double angle formula $sin^2(x) = (1/2)(1 - cos(2x))$ and then a change of formula $u = 2x$ and then apply the Weierstrass substitution.
– DavidG
Nov 21 '18 at 4:20
1
1
You can simplify the algebra by first using the cosine double angle formula $sin^2(x) = (1/2)(1 - cos(2x))$ and then a change of formula $u = 2x$ and then apply the Weierstrass substitution.
– DavidG
Nov 21 '18 at 4:20
You can simplify the algebra by first using the cosine double angle formula $sin^2(x) = (1/2)(1 - cos(2x))$ and then a change of formula $u = 2x$ and then apply the Weierstrass substitution.
– DavidG
Nov 21 '18 at 4:20
add a comment |
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1
$z^4 = (z^2)^2$ so you can use the quadratic formula to get the roots.
– Winther
Nov 21 '18 at 0:19
1
Note that $z^4 - 6z^2 + 1 = (z^2)^2 - 6(z^2) + 1$ can be factored using the quadratic formula. Alternatively, $u = z^2$ works as a $u$-substitution.
– Omnomnomnom
Nov 21 '18 at 0:20
1
You could write $I = int_0^{2pi} frac{1}{1 + frac{1}{2} (1 - cos(2theta))} dtheta = int_0^{2pi} frac{2}{3 - cos(2theta)}dtheta = int_0^{4pi} frac{1}{3 - cos alpha}dalpha = 2 int_0^{2pi} frac{1}{3-cos alpha}dalpha$ and then evaluate that by contour integration.
– Daniel Schepler
Nov 21 '18 at 1:58