Calculating an integral with absolute value

Question

I want to calculate below integration in terms of $a$,$q_1$,$q_2$ and $q_3$ but don't really know how to do.

$ \int_0^{2\pi} |\sin (a-b) \cos (a-b)| \: P \: db$

the form of $P$ is not known but one can use below relations to calculate above integral:

$ \int_0^{2\pi} \sin b \: \sin b \: P \: db = q_1$

$ \int_0^{2\pi} \cos b \: \cos b \: P \: db = q_2$

$ \int_0^{2\pi} \cos b \: \sin b \: P \: db = q_3$

not that the question is asked because I don't know what to do with absolute value. a is a parameter and not a definite value

Is there any really clever guy who could solve this problem? any answers is highly appreciated.

Answer 1

If $$|\sin (a-b) \cos (a-b)|>0,$$

then

\begin{align} I&=\int_0^{2\pi} |\sin (a-b) \cos (a-b)| \: P \: db\\ &=\int_0^{2\pi} \sin (a-b) \cos (a-b) \: P \: db\\ &=\int_0^{2\pi}\{(\sin(a)\cos(b)-\sin(b)\cos(a))(\cos(a)\cos(b)+\sin(a)\sin(b)\}Pdb\\ &=\int_0^{2\pi}\{\sin(a)\cos(a)\cos^2(b)+\cos(b)\sin(b)\sin^2(a)-\sin(b)\cos(b)\cos^2(a)\\&-\cos(a)\sin(a)\sin^2(b)\}Pdb\\ &=\int_0^{2\pi}\{\sin(a)\cos(a)\cos^2(b)+\cos(b)\sin(b)\sin^2(a)-\sin(b)\cos(b)\cos^2(a)\\&-\cos(a)\sin(a)\sin^2(b)\}Pdb\\ &=\sin(a)\cos(a)\int_0^{2\pi}\cos^2(b)Pdb+\sin^2(a)\int_0^{2\pi}\cos(b)\sin(b)Pdb-\cos^2(a)\int\sin(b)\cos(b)Pdb-\cos(a)\sin(a)\int\sin^2(b)Pdb\\ &=\sin(a)\cos(a)q_2+\sin^2(a)q_3-\cos^2(a)q_3-\cos(a)\sin(a)q_1\\ &=\frac{\sin(2a)}{2}\left(q_2-q_1\right)-\cos(2a)q_3 \end{align}

If $$|\sin (a-b) \cos (a-b)|<0,$$

then we have an overall negative sign.