integral of 1/(sin^2theta)+1 integrated from -pi to pi
just give me the steps on how to go about it i don't want you to do it for me but i just don't know how to start it and the answer as well so i know i have done it right afterwards thank you
just give me the steps on how to go about it i don't want you to do it for me but i just don't know how to start it and the answer as well so i know i have done it right afterwards thank you
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I'll highlight the main points.
Rewrite in terms of the complex exponential:
∫(θ = -π to π) dθ/(sin^2(θ) + 1)
= ∫(θ = -π to π) dθ/[((e^(iθ) - e^(-iθ))/(2i))^2 + 1]
= ∫(θ = -π to π) -4 dθ/[e^(2iθ) + e^(-2iθ) - 6].
Now, write this as a contour integral with z = e^(iθ) with θ in [-π, π],
which traces out the unit circle:
Since z = e^(iθ), 1/z = e^(-iθ), and dz = ie^(iθ) dθ ==> dz/(iz) = dθ,
the integral transforms to
∫c -4 (dz/(iz)) / (z^2 + 1/z^2 - 6)
= ∫c 4iz dz / (z^4 - 6z^2 + 1).
Now, use the Residue Theorem to evaluate this integral
(after locating the singularities which lie inside |z| = 1).
Singularities (all of which are simple poles):
z^4 - 6z^2 + 1 = 0
==> (z^2 - 3)^2 = 8
==> z^2 = 3 ± 2√2 = (1 ± √2)^2
==> z = ±(1 + √2), or ±(1 - √2); only the latter pair is in |z| = 1.
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Compute the residues at z = ±(1 - √2):
lim(z → ±(1 - √2)) (z - ±(1 - √2)) * 4iz/(z^4 - 6z^2 + 1)
= 4i * lim(z → ±(1 - √2)) (z^2 - ±(1 - √2)z) / (z^4 - 6z^2 + 1)
= 4i * lim(z → ±(1 - √2)) (2z - ±(1 - √2)) / (4z^3 - 12z), via L'Hopital's Rule
= i * lim(z → ±(1 - √2)) (2z - ±(1 - √2)) / (z(z^2 - 3))
= i * ±(1 - √2) / (±(1 - √2) * ((3 - 2√2) - 3))
= -i/(2√2).
Hence, the integral equals (by the Residue Theorem)
2πi * [-i/(2√2) + -i/(2√2)] = π√2.
Double check:
http://www.wolframalpha.com/input/?i=%E2…
I hope this helps!
Rewrite in terms of the complex exponential:
∫(θ = -π to π) dθ/(sin^2(θ) + 1)
= ∫(θ = -π to π) dθ/[((e^(iθ) - e^(-iθ))/(2i))^2 + 1]
= ∫(θ = -π to π) -4 dθ/[e^(2iθ) + e^(-2iθ) - 6].
Now, write this as a contour integral with z = e^(iθ) with θ in [-π, π],
which traces out the unit circle:
Since z = e^(iθ), 1/z = e^(-iθ), and dz = ie^(iθ) dθ ==> dz/(iz) = dθ,
the integral transforms to
∫c -4 (dz/(iz)) / (z^2 + 1/z^2 - 6)
= ∫c 4iz dz / (z^4 - 6z^2 + 1).
Now, use the Residue Theorem to evaluate this integral
(after locating the singularities which lie inside |z| = 1).
Singularities (all of which are simple poles):
z^4 - 6z^2 + 1 = 0
==> (z^2 - 3)^2 = 8
==> z^2 = 3 ± 2√2 = (1 ± √2)^2
==> z = ±(1 + √2), or ±(1 - √2); only the latter pair is in |z| = 1.
-------------------
Compute the residues at z = ±(1 - √2):
lim(z → ±(1 - √2)) (z - ±(1 - √2)) * 4iz/(z^4 - 6z^2 + 1)
= 4i * lim(z → ±(1 - √2)) (z^2 - ±(1 - √2)z) / (z^4 - 6z^2 + 1)
= 4i * lim(z → ±(1 - √2)) (2z - ±(1 - √2)) / (4z^3 - 12z), via L'Hopital's Rule
= i * lim(z → ±(1 - √2)) (2z - ±(1 - √2)) / (z(z^2 - 3))
= i * ±(1 - √2) / (±(1 - √2) * ((3 - 2√2) - 3))
= -i/(2√2).
Hence, the integral equals (by the Residue Theorem)
2πi * [-i/(2√2) + -i/(2√2)] = π√2.
Double check:
http://www.wolframalpha.com/input/?i=%E2…
I hope this helps!
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