The value of

\underset{0}{\overset{\frac{\pi }{2}}{\int }}|\sin x+\sin 2x+\sin 3x|dx\mathrm{<br>}    is

To find the value of the given definite integral, we start by analyzing the integrand:

$I={\int }_0^{\frac{\pi }{2}}|\sin x+\sin 2x+\sin 3x|dx$

Step 1: Simplify the trigonometric expression inside the absolute value
Let us define the function $f\left(x\right)=\sin x+\sin 2x+\sin 3x$. We can group the first and the third terms and apply the sum-to-product formula $\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$:

$\sin x+\sin 3x=2\sin 2x\cos x$

Substituting this back into the function gives:

$f\left(x\right)=2\sin 2x\cos x+\sin 2x$

Factoring out $\sin 2x$:

$f\left(x\right)=\sin 2x(2\cos x+1)$

Step 2: Determine the sign of the integrand in the interval of integration
The integration is carried out over the interval $[0,\frac{\pi }{2}]$. Let's examine the sign of each factor in this interval:
1. For $x\in [0,\frac{\pi }{2}]$, we have $2x\in [0,\pi ]$. In this range, the sine function is non-negative, so $\sin 2x\ge 0$.
2. For $x\in [0,\frac{\pi }{2}]$, the cosine function is also non-negative ($\cos x\ge 0$), which means $2\cos x+1\ge 1>0$.

Since both factors are non-negative on the interval $[0,\frac{\pi }{2}]$, the product is also non-negative: $f\left(x\right)\ge 0$.
Therefore, we can remove the absolute value signs:

$|\sin x+\sin 2x+\sin 3x|=\sin x+\sin 2x+\sin 3x$

Step 3: Evaluate the definite integral
Now, we compute the integral directly by finding the antiderivative of each term:

$I={\int }_0^{\frac{\pi }{2}}(\sin x+\sin 2x+\sin 3x)dx$

Using the integration rule $\int \sin kxdx=-\frac{\cos kx}{k}$, we obtain:

$I={[-\cos x-\frac{\cos 2x}{2}-\frac{\cos 3x}{3}]}_0^{\frac{\pi }{2}}$

Now we evaluate this expression at the upper limit $x=\frac{\pi }{2}$ and subtract the value at the lower limit $x=0$:

At the upper limit $x=\frac{\pi }{2}$:
$-\cos \left(\frac{\pi }{2}\right)-\frac{\cos \pi }{2}-\frac{\cos \left(\frac{3\pi }{2}\right)}{3}=-0-\frac{-1}{2}-0=\frac{1}{2}$

At the lower limit $x=0$:
$-\cos 0-\frac{\cos 0}{2}-\frac{\cos 0}{3}=-1-\frac{1}{2}-\frac{1}{3}=-\frac{6+3+2}{6}=-\frac{11}{6}$

Subtracting the lower limit evaluation from the upper limit evaluation:

$I=\frac{1}{2}-(-\frac{11}{6})=\frac{3}{6}+\frac{11}{6}=\frac{14}{6}=\frac{7}{3}$

Thus, the value of the integral is 7/3.