Calculate the integral: integral from - pi to () pi x(2+

Calculate the integral: $\int_{-\pi}^\pi x(2+\cos x) \cos^{17} x \, dx$.

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

Step 1: Expand the integrand

\[\int_{-\pi}^\pi x(2+\cos x) \cos^{17} x \, dx = \int_{-\pi}^\pi (2x\cos^{17} x + x\cos x \cdot \cos^{17} x) \, dx\]

\[= 2\int_{-\pi}^\pi x\cos^{17} x \, dx + \int_{-\pi}^\pi x\cos^{18} x \, dx\]

Step 2: Analyze the symmetry of each integrand

For the first integral $\int_{-\pi}^\pi x\cos^{17} x \, dx$: Let $f(x) = x\cos^{17} x$. Since $\cos(-x)=\cos(x)$:

\[f(-x) = (-x)[\cos(-x)]^{17} = (-x)[\cos x]^{17} = -x\cos^{17} x = -f(x)\]

Since $f(-x)=-f(x)$, the function $x\cos^{17}x$ is odd, so $\int_{-\pi}^\pi x\cos^{17}x\,dx = 0$.

For the second integral $\int_{-\pi}^\pi x\cos^{18} x \, dx$: Let $g(x)=x\cos^{18}x$. Similarly:

\[g(-x) = (-x)[\cos(-x)]^{18} = -x\cos^{18}x = -g(x)\]

Since $g(-x)=-g(x)$, the function $x\cos^{18}x$ is also odd, so $\int_{-\pi}^\pi x\cos^{18}x\,dx = 0$.

Step 3: Combine the results

\[\int_{-\pi}^\pi x(2+\cos x) \cos^{17} x \, dx = 2\int_{-\pi}^\pi x\cos^{17} x \, dx + \int_{-\pi}^\pi x\cos^{18} x \, dx = 2 \cdot 0 + 0 = 0\]

General principle:

For any function of the form $f(x) = x \cdot p(\cos x)$ where $p$ is any polynomial or function of $\cos x$: since $x$ is odd and $\cos x$ is even, any power $\cos^n x$ is also even. The product of an odd function ($x$) and an even function ($p(\cos x)$) is always an odd function, so $\int_{-a}^a x \cdot p(\cos x) \, dx = 0$ for any symmetric interval $[-a, a]$.

Alternative verification:

We can verify the oddness directly:

\[f(x) = x(2+\cos x)\cos^{17} x, \qquad f(-x) = (-x)(2+\cos(-x))\cos^{17}(-x) = (-x)(2+\cos x)\cos^{17} x = -f(x)\]

Since $f$ is odd and the interval $[-\pi, \pi]$ is symmetric about the origin, the integral equals zero.

Geometric interpretation:

The function $f(x) = x(2+\cos x)\cos^{17} x$ is odd, meaning its graph is symmetric about the origin. When we integrate over a symmetric interval $[-\pi, \pi]$, the negative area on $[-\pi, 0]$ exactly cancels the positive area on $[0, \pi]$, resulting in a total integral of zero.

Final answer:

\[\boxed{\int_{-\pi}^\pi x(2+\cos x) \cos^{17} x \, dx = 0}\]