Solution:

1) $\int_0^{2\pi} \frac{d\theta}{5+3sin\theta}$

Let $e^{i\theta} =z$-------(1)

diff.(1) w.r.t $\theta$

$e^{i\theta}.id\theta =dz$

$z.id\theta=dz$

$d\theta=\frac{dz}{iz}$-------(2)

$sin\theta =\frac{z^2-1}{2iz}$-----(3)

$I=\int_c \frac {1}{5+3(\frac{z^2-1}{2iz})}\frac{dz}{iz}$

$I=\int_c \frac {2iz}{3z^2-3+10iz}\frac{dz}{iz}$

$I=\int_c \frac {2}{3z^2+10iz-3}{dz}$

$I=\int_c \frac {2}{(3z+1)(z+3i)}{dz}$

where c is the circle |z|=1

Now the poles are ,

3z+i=0 & z+3i

$z=\frac{-i}{3}$ and $ z=-3i $

An poles $z=\frac{-i}{3}$ lies inside the circle, and pole $z= -3i$ lies outside the circle.

Residue at $z=\frac{-i}{3}$

$\lim_{z\to -1/3}[z+\frac {i}{3}]\frac{2}{(3z+i)(z+3i)}$

$\lim_{z\to -1/3}[z+\frac {i}{3}]\frac{2}{3(z+\frac{i}{3})(z+3i)}$

$\lim_{z\to -1/3}\frac{2}{3(z+3i)}$

$\therefore \frac{2}{3(\frac{-i}{3}+3i)}$

=$\frac{2}{8i}$

=$\frac{1}{4i}$

$I=2\pi i \times \frac{1}{4i}$

$I =\frac{\pi}{2}$

2) $\int_{-\infty} ^\infty \frac {x^2}{(x^2+a^2)(x^2+b^2)}dx$

i) $f(x)=\frac{x^2}{(x^2+a^2)(x^2+b^2)}$

$f(z)=\frac{z^2}{(z^2+a^2)(z^2+b^2)}$

2)$zf(z)=z\frac{z^2}{(z^2+a^2)(z^2+b^2)}$

$zf(z)=\frac{z^3}{(z^2+a^2)(z^2+b^2)}$

3)Now the poles are

z= +ai, -ai and z=+bi, z=-bi

where poles z= ai and z=bi lies in the upper half of the z plane

4)residue at z= ai,

=$\lim{z \to ai} (z-ai)\frac {z^2}{(z-ai)(z+ai)(z^2+b^2)}$

=$\lim{z \to ai} \frac {z^2}{(z+ai)(z^2+b^2)}$

=$\frac {ai^2}{(ai+ai)((ai)^2+b^2)}$

=$\frac {a^2}{2ai(-a^2+b^2)}$

Residue at z= $ai = \frac{a}{2i(a^2-b^2)}$-------(1)

similary residue at z=bi is

$\lim{z \to bi} (z-bi)\frac{z^2}{(z^2+a^2)(z+bi)(z-bi)}$

$\lim{z \to bi} \frac{z^2}{(z^2+a^2)(z+bi)}$

$\frac{(bi)^2}{[(bi^2)+a^2][bi+bi]}$

$\frac{-b^2}{2bi[a^2-b^2]}$

Residue at z= -bi = $\frac{-b}{2i[a^2-b^2]}$

sum of residue = [residue at z=-ai+residue at z=-bi]

$2\pi i[\frac{a}{2i(a^2-b^2)}-\frac{b}{2i(a^2-b^2)}]$

5)$\int_{-\infty} ^\infty \frac {x^2}{(x^2+a^2)(x^2+b^2)}dx$

=2$\pi i (sum \ of \ residue)$

=2$\pi i[\frac {a}{2i(a^2-b^2)}-\frac {b}{2i(a^2-b^2)}]$

=2$\pi i[\frac {a-b}{2i(a^2-b^2)}]$

=$\frac{2\pi i}{2i} [\frac {a-b}{(a+b)(a-b)}]$

=$\frac{\pi}{(a+b)}$

$\int_{-\infty} ^\infty \frac {x^2}{(x^2+a^2)(x^2+b^2)}dx$ =$\frac{\pi}{(a+b)}$