$$6\int_{0}^{\pi} |(\sin3x+\sin2x+\sin x)|dx$$ is equal to _________.

given
$$6\int_0^{\pi}\ |\sin3x+\sin2x+\sin x|dx$$

sin 3x + sin x = 2sin 2x cos x

So the integrand becomes:
sin 2x(2cos x + 1)
Analysing the sign of modulus by finding zeros

$$(\sin2x=0\Rightarrow x=0,\frac{\pi}{2},\pi)\ (2\cos x+1=0\Rightarrow x=\frac{2\pi}{3})$$

Split intervals:

• ((0, $$\frac{\pi}{2}$$)): positive
• (($$\frac{\pi}{2}$$, $$\frac{2\pi}{3}$$)): negative
• (($$\frac{2\pi}{3}$$,$$\pi$$)): positive

So:
$$\int_0^{\pi}|f(x)|dx=\int_0^{\pi/2}f(x)dx-\int_{\pi/2}^{2\pi/3}f(x)dx+\int_{2\pi/3}^{\pi}f(x)dx$$

Let:
$$F(x)=-\frac{4}{3}\cos^3x-\cos^2x$$

Evaluate:

• ($$\int_0^{\pi/2}$$ = $$\frac{7}{3}$$)
• ($$\int_{\pi/2}^{2\pi/3}$$ = -$$\frac{1}{12}$$)
• ($$\int_{2\pi/3}^{\pi}$$ =$$\frac{5}{12}$$)

So total on calculation

$$\int_0^{\pi}|f(x)|dx=\frac{7}{3}+\frac{1}{12}+\frac{5}{12}=\frac{17}{6}$$

6 $$\frac{17}{6}$$ = 17 

17 is the answer