1) Prove the equality integral from 0 to () pi x f(sin x)

1) Prove the equality $\int_0^\pi x f(\sin x) \, dx = \frac{\pi}{2} \int_0^\pi f(\sin x) \, dx$, where $f$ is continuous on $[0, \pi]$. 2) Calculate $\int_0^\pi x \frac{\sin x}{1+\cos^2 x} \, dx$.

Part 1: Prove $\int_0^\pi x f(\sin x) \, dx = \frac{\pi}{2} \int_0^\pi f(\sin x) \, dx$

Proof:

Let $I = \int_0^\pi x f(\sin x) \, dx$. We use the substitution $u = \pi - x$, so $du = -dx$, $dx=-du$, $x=\pi-u$.

Change the limits of integration:

When $x = 0$: $u = \pi$. When $x = \pi$: $u = 0$.

Key observation about sine:

\[\sin x = \sin(\pi - u) = \sin u\]

Transform the integral:

\[I = \int_0^\pi x f(\sin x) \, dx = \int_\pi^0 (\pi - u) f(\sin u) \cdot (-du) = \int_0^\pi (\pi - u) f(\sin u) \, du\]

Since the variable of integration is a dummy variable, we can replace $u$ with $x$:

\[I = \int_0^\pi (\pi - x) f(\sin x) \, dx = \pi \int_0^\pi f(\sin x) \, dx - I\]

Solve for $I$:

\[2I = \pi \int_0^\pi f(\sin x) \, dx \quad\Rightarrow\quad I = \frac{\pi}{2} \int_0^\pi f(\sin x) \, dx\]

Therefore:

\[\boxed{\int_0^\pi x f(\sin x) \, dx = \frac{\pi}{2} \int_0^\pi f(\sin x) \, dx}\]

Geometric interpretation:

This shows that weighting $f(\sin x)$ by $x$ over $[0, \pi]$ gives the same result as weighting it uniformly by $\frac{\pi}{2}$ (the interval's midpoint). This is because the symmetry of $\sin x$ about $x = \frac{\pi}{2}$ on $[0, \pi]$ creates a balance: extra weight on the right side ($x>\frac\pi2$) exactly compensates for reduced weight on the left side ($x<\frac\pi2$).

Part 2: Calculate $\int_0^\pi x \frac{\sin x}{1+\cos^2 x} \, dx$

We apply the result from Part 1 with $f(\sin x) = \frac{\sin x}{1+\cos^2 x}$, noting $f$ is continuous on $[0, \pi]$ since $1+\cos^2 x \geq 1 > 0$. From Part 1:

\[\int_0^\pi x \frac{\sin x}{1+\cos^2 x} \, dx = \frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\cos^2 x} \, dx\]

Evaluate $\int_0^\pi \frac{\sin x}{1+\cos^2 x} \, dx$ using substitution:

Let $u = \cos x$, so $du = -\sin x \, dx$, i.e. $\sin x \, dx = -du$. When $x=0$: $u=1$; when $x=\pi$: $u=-1$.

\[\int_0^\pi \frac{\sin x}{1+\cos^2 x} \, dx = \int_1^{-1} \frac{-du}{1+u^2} = \int_{-1}^1 \frac{du}{1+u^2} = \left[\arctan u\right]_{-1}^1 = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}\]

Final calculation:

\[\int_0^\pi x \frac{\sin x}{1+\cos^2 x} \, dx = \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^2}{4}\]

Verification of the result:

The integrand $x \frac{\sin x}{1+\cos^2 x}$ is positive on $(0, \pi)$; the maximum of $x$ on $[0,\pi]$ is $\pi$, the maximum of $\sin x$ is $1$, and the minimum of $1+\cos^2 x$ is $1$ (at $\cos x=0$). So the integrand is bounded, and $\frac{\pi^2}{4}\approx2.47$ is a reasonable value.

General principle:

The result from Part 1 is a powerful tool: for any continuous $f$, $\int_0^\pi x f(\sin x) \, dx = \frac{\pi}{2} \int_0^\pi f(\sin x) \, dx$, useful for simplifying many integrals involving $x$ and $\sin x$ over $[0, \pi]$.

Final answer:

\[\boxed{\int_0^\pi x \frac{\sin x}{1+\cos^2 x} \, dx = \frac{\pi^2}{4}}\]