How to find $\int_0^{2\pi} |a\cos(x)+b\sin(x)|dx$, where $a^2+b^2=1$ I need help to solve this integral:
$$\int_0^{2\pi} |a\cos(x)+b\sin(x)|dx$$  where $a^2+b^2=1$. I hope someone is able to help me.
 A: Choose $\theta$ such that $\sin(\theta) = a$ and $\cos(\theta)=b$. Then you have $a\cos(x)+b\sin(x) = \sin(\theta)\cos(x)+\cos(\theta)\sin(x)$. A standard trig. identity shows that this can be written as $a\cos(x)+b\sin(x) = \sin(\theta+x)$. Since $x \mapsto \sin(\theta+x)$ is $2 \pi$ periodic, then the integral can be written as $\int_0^{2 \pi} |\sin(\theta+x)|dx = \int_0^{2 \pi} |\sin(x)|dx$ = $\int_0^{\pi} \sin(x)dx - \int_{\pi}^{2\pi} \sin(x)dx = 2 \int_0^{\pi} \sin(x)dx$. I trust you can take it from here.
A: Hint: Let $\phi$ be a number such that $a=\sin\phi$ and $b=\cos\phi$. You want to integrate $|\sin(x+\phi)|$. Maybe make the natural change of variable, or let the geometry guide you to the answer.
Remark: The "trick" above has a number of uses. The idea works even if $a^2+b^2\ne 1$. For suppose that $a^2+b^2\ne 0$. Note that 
$$a\cos x+b\sin x=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x\right).$$
Let $\phi$ be a number such that $a/\sqrt{a^2+b^2}=\sin \phi$ and $b/\sqrt{a^2+b^2}=\cos\phi$.
