I have two integrals that I need to solve. $q$ and $k$ are both arbitrary positive real numbers.
$$\int_0^\pi{\frac{\sin{\theta}}{(1+(k^2+q^2-2kq\cos{\theta})a^2)^2}}d\theta$$ $$\int_0^\pi{\frac{\sin{\theta}}{(1+(k^2+q^2+2kq\cos{\theta})a^2)^4}}d\theta $$
I don't really have any good ideas on how to approach this, but when I tried using Mathematica I got the following results $$\frac{2}{1+a^4(k^2-q^2)^2+2a^2(k^2+q^2)}$$
$$\frac{1}{6a^2kq}\Big(\frac{1}{(1+a^2(k-q)^2)^3}-\frac{1}{(1+a^2(k+q)^2)^3}\Big)$$ for the first and second integral respectively, but with a bunch of additional conditions so I would prefer to solve the integral using pen and paper to be sure about the correctness of all the intermediary steps.
Maybe the Mathematica solutions can give a hint on how to approach the manual solution. Any tips even just on how to start or approach this would be greatly appreciated. Thank you!
Start with the integral \begin{align} I(x)=&\int_0^\pi{\frac{\sin{\theta}}{x+(k^2+q^2-2kq\cos{\theta})a^2}}d\theta =\frac1{2kqa^2}\ln \frac{x+(k+q)^2a^2}{x+(k-q)^2a^2} \end{align} Then
\begin{align} &\int_0^\pi{\frac{\sin{\theta}}{(1+(k^2+q^2-2kq\cos{\theta})a^2)^2}}d\theta\\ =&-\frac{d I(x)}{dx}\bigg|_{x=1}=\frac{2}{1+a^4(k^2-q^2)^2+2a^2(k^2+q^2)} \end{align} and
\begin{align} &\int_0^\pi{\frac{\sin{\theta}}{(1+(k^2+q^2+2kq\cos{\theta})a^2)^4}}d\theta\\ =&-\frac16 \frac{d^3 I(x)}{dx^3}\bigg|_{x=1}=\frac{1}{6a^2kq}\Big(\frac{1}{(1+a^2(k-q)^2)^3}-\frac{1}{(1+a^2(k+q)^2)^3}\Big) \end{align}