Section 6.2, Question 1: By the identity sin theta = cos pi/2 - theta and the change of variables alpha = pi/2 -
theta, the integral is equal to the integral of 1/(2 - cos alpha) from -3pi/2 to pi/2, which is the same as the
integral from 0 to 2 pi, which is 2 pi/sqrt(3) by Example 2 of Section 6.2. (Yes, you could do it by residue
calculus, and you should, but sometimes you can cheat too).
Question 4: We can use residue calculus immediately, but we can save some effort by noting that 1 + sin^2
theta = (1 + (1 - cos 2 theta)/2) = (3 - cos 2 theta)/2, and rewriting the integral as
int_{-pi}^{pi} 2 dtheta / (3 - cos 2theta) = int_{-2pi}^{2pi}
dalpha / (3 - cos alpha).
Using the change of variables z = e^{i alpha} we get
int_Gamma (dz/iz) / (3 - (z + 1/z)/2) = 2i int_Gamma dz/(z^2 - 6z + 1)
where Gamma is the unit circle traversed
twice
anticlockwise.
The intgrand has poles at 3+sqrt(8) and 3-sqrt(8); the pole at 3+sqrt(8)
is outside the contour and is therefore irrelevant.
At 3-sqrt(8) the
residue is -1/(2sqrt(8)), so the integral as a whole is -4pi i /(2 sqrt(8)),
(remember that the contour winds around the pole twice), so the final answer
is 2i (-4 pi i/(2 sqrt(8))) = pi sqrt(2).
Question 5: We make the usual change of variables z = e^{i theta} and
rewrite the integral as
int_Gamma (dz/iz) 1/(1 + a (z + 1/z)/2) = 2i int_Gamma dz / (az^2 + 2z
+ a).
This has simple poles at (-1 + sqrt(1-a^2))/a and (-1 - sqrt(1-a^2))/a.
The latter pole is obviously outside the unit circle, since -1 -sqrt(1-a^2)
is outside the circle and 1/a has magnitude > 1.
The former pole is
the reciprocal of the latter pole and is therefore inside the unit circle.
The residue at this pole is -1/[2sqrt(1-a^2)], so the integral is
-2pi i/[2 sqrt(1 - a^2)], and the final answer is 2i [-2pi i/[2 sqrt(1 - a^2)]]
= 2 pi / sqrt(1 - a^2).
Section 6.3, Question 1.
We compute the integral of dz/(z^2 + 2z + 2)
from -R to R.
We may add and subtract the contour C^R_+, the upper
anticlockwise semicircular contour of radius R.
When you add C^R_+,
the contour becomes closed, and we can use residue theory to work out
the integral; the poles are at i-1 and -i-1, and the residue at i-1 (which
is the only one that we care about) is 1/(2i), so the integral is 2pi i/2i
= pi.
We still have the error term to deal with, which is negative
the integral over C^R_+, but that goes to zero by Lemma 1 of Section 6.3.