$\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|dx$ $\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|dx$
MyAttempt
$\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|=\int_{0}^{\pi/2}\sqrt2\sin x+2\cos x dx+\int_{\pi/2}^{\pi}|\sqrt2\sin x+2\cos x| dx$
I could solve first integral but in second one,i could not judge the mod will take plus sign or minus sign or how to break it in further intervals?

We Can Write $\displaystyle \sqrt{2}\sin x+2\cos x = \sqrt{\left(\sqrt{2}\right)^2+2^2}\left\{\frac{\sqrt{2}}{\sqrt{6}}\sin x+\frac{2}{\sqrt{6}}\cos x\right\}$
$\displaystyle  = \sqrt{6}\left\{\sin x\cdot \frac{1}{\sqrt{3}}+\cos x\cdot \frac{2}{\sqrt{6}}\right\} = \sqrt{6}\left\{\sin x\cdot \cos \alpha+\cos x\cdot \sin \alpha\right\} = \sqrt{6}\cdot \sin (x+\alpha)$
Where $$\displaystyle \sin \alpha = \frac{2}{\sqrt{6}}$$ and $$\displaystyle \cos \alpha = \frac{1}{\sqrt{3}}$$ and $$\displaystyle \tan \alpha = \sqrt{2}$$ . So $\displaystyle \frac{\pi}{4}<\alpha <\frac{\pi}{3}$
So Integral Convert into $$\displaystyle \sqrt{6}\int_{0}^{\pi}\left|\sin (x+\alpha)\right|dx$$
Let $(x+\alpha) = t\;,$ Then $dx = dt$ and Changing Limit, we get
$$\displaystyle I = \sqrt{6}\int_{\alpha}^{\pi+\alpha}|\sin t|dt = \sqrt{6}\int_{\alpha}^{\pi}\sin tdt-\sqrt{6}\int_{\pi}^{\pi+\alpha}\sin tdt $$
So we get $$\displaystyle I = \sqrt{6}\left(1+\cos \alpha\right)+\sqrt{6}\left(-\cos \alpha +1\right) = 2\sqrt{6}$$
