Calculate the integral: integral from - pi to () pi x(sin

Calculate the integral: $\int_{-\pi}^\pi x(\sin x-1) \, dx$.

We will use properties of even and odd functions to simplify the calculation.

Step 1: Expand the integrand

\[\int_{-\pi}^\pi x(\sin x-1) \, dx = \int_{-\pi}^\pi (x\sin x - x) \, dx = \int_{-\pi}^\pi x\sin x \, dx - \int_{-\pi}^\pi x \, dx\]

Step 2: Analyze the symmetry of each integrand

For the first integral $\int_{-\pi}^\pi x\sin x \, dx$: Let $f(x) = x\sin x$. Check parity:

\[f(-x) = (-x)\sin(-x) = (-x)(-\sin x) = x\sin x = f(x)\]

Since $f(-x) = f(x)$, the function $x\sin x$ is even. For an even function over $[-a,a]$: $\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx$.

For the second integral $\int_{-\pi}^\pi x \, dx$: Let $g(x) = x$. Since $g(-x)=-x=-g(x)$, the function $x$ is odd, so $\int_{-a}^a g(x)\,dx=0$.

Step 3: Simplify using symmetry

\[\int_{-\pi}^\pi x(\sin x-1) \, dx = \int_{-\pi}^\pi x\sin x \, dx - \int_{-\pi}^\pi x \, dx = 2\int_0^\pi x\sin x \, dx - 0 = 2\int_0^\pi x\sin x \, dx\]

Step 4: Evaluate $\int_0^\pi x\sin x \, dx$ using integration by parts

Let $u = x$ and $dv = \sin x \, dx$, so $du = dx$ and $v = -\cos x$.

\[\int_0^\pi x\sin x \, dx = \left[-x\cos x\right]_0^\pi - \int_0^\pi (-\cos x) \, dx = \left[-x\cos x\right]_0^\pi + \left[\sin x\right]_0^\pi\]

Step 5: Evaluate the boundary terms

For $\left[-x\cos x\right]_0^\pi$: at $x=\pi$, $-\pi\cos(\pi)=-\pi(-1)=\pi$; at $x=0$, $-0\cos(0)=0$. So $\left[-x\cos x\right]_0^\pi = \pi$.

For $\left[\sin x\right]_0^\pi$: at $x=\pi$, $\sin(\pi)=0$; at $x=0$, $\sin(0)=0$. So $\left[\sin x\right]_0^\pi = 0$.

Step 6: Complete the calculation

\[\int_0^\pi x\sin x \, dx = \pi + 0 = \pi\]

Therefore:

\[\int_{-\pi}^\pi x(\sin x-1) \, dx = 2\int_0^\pi x\sin x \, dx = 2\pi\]

Verification using direct computation:

We could also verify that $\int_{-\pi}^\pi x \, dx = 0$ directly:

\[\int_{-\pi}^\pi x \, dx = \left[\frac{x^2}{2}\right]_{-\pi}^\pi = \frac{\pi^2}{2} - \frac{(-\pi)^2}{2} = 0\]

Geometric interpretation:

The integral splits into two parts: $\int_{-\pi}^\pi x\sin x \, dx$ is the signed area under the even function $y=x\sin x$, and by symmetry the positive area on $[0,\pi]$ equals that on $[-\pi,0]$, giving total $2\pi$; while $\int_{-\pi}^\pi x \, dx$ is the signed area under the odd function $y=x$, where the negative area on $[-\pi,0]$ exactly cancels the positive area on $[0,\pi]$, giving total $0$.

Final answer:

\[\boxed{\int_{-\pi}^\pi x(\sin x-1) \, dx = 2\pi}\]