Key Concepts and Formulas

• King's Property: For a definite integral $\int_0^a f(x) dx$, we have $\int_0^a f(x) dx = \int_0^a f(a-x) dx$.
• Integral Property: For a definite integral $\int_0^{2a} f(x) dx$, if $f(2a-x) = -f(x)$, then $\int_0^{2a} f(x) dx = 0$.
• Trigonometric Identities:
  • $\sin^2 x + \cos^2 x = 1$
  • $\sin(2x) = 2 \sin x \cos x$
  • $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = 1 - \frac{1}{2} \sin^2(2x)$

Step-by-Step Solution

Let the given integral be $I$. I = \int_\limits0^\pi \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x

Step 1: Apply the King's Property. We use the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$. Here, $a = \pi$. Let $f(x) = \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x}$. Then $f(\pi - x) = \frac{(\pi - x)^2 \sin(\pi - x) \cos(\pi - x)}{\sin^4(\pi - x) + \cos^4(\pi - x)}$. We know that $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$. So, $\sin(\pi - x) \cos(\pi - x) = (\sin x)(-\cos x) = -\sin x \cos x$. And $\sin^4(\pi - x) = (\sin x)^4 = \sin^4 x$, and $\cos^4(\pi - x) = (-\cos x)^4 = \cos^4 x$. Therefore, $f(\pi - x) = \frac{(\pi - x)^2 (-\sin x \cos x)}{\sin^4 x + \cos^4 x}$. Applying the King's property: $I = \int_0^\pi \frac{(\pi - x)^2 \sin(\pi - x) \cos(\pi - x)}{\sin^4(\pi - x) + \cos^4(\pi - x)} dx = \int_0^\pi \frac{(\pi - x)^2 (-\sin x \cos x)}{\sin^4 x + \cos^4 x} dx$ So, we have two expressions for $I$: $I = \int_0^\pi \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \quad (*)$ $I = \int_0^\pi \frac{(\pi - x)^2 (-\sin x \cos x)}{\sin ^4 x+\cos ^4 x} d x \quad (**)$

Step 2: Add the two expressions for I. Adding (**) to (*): $2I = \int_0^\pi \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx + \int_0^\pi \frac{(\pi - x)^2 (-\sin x \cos x)}{\sin ^4 x+\cos ^4 x} dx$ $2I = \int_0^\pi \frac{[x^2 - (\pi - x)^2] \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$ Simplify the term $x^2 - (\pi - x)^2$: $x^2 - (\pi - x)^2 = x^2 - (\pi^2 - 2\pi x + x^2) = x^2 - \pi^2 + 2\pi x - x^2 = 2\pi x - \pi^2 = \pi(2x - \pi)$. So, $2I = \int_0^\pi \frac{\pi(2x - \pi) \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$ $2I = \pi \int_0^\pi \frac{(2x - \pi) \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$

Step 3: Consider the integral $\int_0^\pi g(x) dx$ where $g(x) = \frac{(2x - \pi) \sin x \cos x}{\sin ^4 x+\cos ^4 x}$. Let's check the property $g(\pi - x) = -g(x)$. We already found that $\sin(\pi - x) \cos(\pi - x) = -\sin x \cos x$. And $\sin^4(\pi - x) + \cos^4(\pi - x) = \sin^4 x + \cos^4 x$. The term $(2x - \pi)$ becomes $(2(\pi - x) - \pi) = (2\pi - 2x - \pi) = \pi - 2x = -(2x - \pi)$. So, $g(\pi - x) = \frac{(2(\pi - x) - \pi) \sin(\pi - x) \cos(\pi - x)}{\sin^4(\pi - x) + \cos^4(\pi - x)}$ $g(\pi - x) = \frac{-(2x - \pi) (-\sin x \cos x)}{\sin^4 x + \cos^4 x} = \frac{(2x - \pi) \sin x \cos x}{\sin^4 x + \cos^4 x} = g(x)$.

This is not $-g(x)$. Let's re-examine the addition of $I$ and the modified $I$.

Let's go back to $2I = \int_0^\pi \frac{[x^2 - (\pi - x)^2] \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$. $2I = \int_0^\pi \frac{(2\pi x - \pi^2) \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$ $2I = \int_0^\pi \frac{2\pi x \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx - \int_0^\pi \frac{\pi^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$

Let $I_1 = \int_0^\pi \frac{2\pi x \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$ and $I_2 = \int_0^\pi \frac{\pi^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$. So, $2I = I_1 - I_2$.

For $I_1$: Let $h(x) = \frac{2\pi x \sin x \cos x}{\sin ^4 x+\cos ^4 x}$. $h(\pi - x) = \frac{2\pi (\pi - x) \sin(\pi - x) \cos(\pi - x)}{\sin^4(\pi - x) + \cos^4(\pi - x)}$ $h(\pi - x) = \frac{2\pi (\pi - x) (-\sin x \cos x)}{\sin^4 x + \cos^4 x} = \frac{-2\pi (\pi - x) \sin x \cos x}{\sin^4 x + \cos^4 x}$. So, $I_1 = \int_0^\pi h(x) dx$. Also, $I_1 = \int_0^\pi h(\pi - x) dx = \int_0^\pi \frac{-2\pi (\pi - x) \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. Adding the two forms of $I_1$: $2I_1 = \int_0^\pi \frac{2\pi x \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx + \int_0^\pi \frac{-2\pi (\pi - x) \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$ $2I_1 = \int_0^\pi \frac{[2\pi x - 2\pi (\pi - x)] \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$ $2I_1 = \int_0^\pi \frac{(2\pi x - 2\pi^2 + 2\pi x) \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$ $2I_1 = \int_0^\pi \frac{(4\pi x - 2\pi^2) \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$. This is not simplifying as expected.

Let's use a different approach for the integral I = \int_\limits0^\pi \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x. We can rewrite the denominator: $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = 1 - \frac{1}{2} (2 \sin x \cos x)^2 = 1 - \frac{1}{2} \sin^2(2x)$. The numerator is $x^2 \sin x \cos x = x^2 \frac{\sin(2x)}{2}$. So, $I = \int_0^\pi \frac{x^2 \frac{\sin(2x)}{2}}{1 - \frac{1}{2} \sin^2(2x)} dx = \int_0^\pi \frac{x^2 \sin(2x)}{2 - \sin^2(2x)} dx$.

Now, let's use the property $\int_0^{2a} f(x) dx = \int_0^{2a} f(2a-x) dx$. Here, the integral is from $0$ to $\pi$. Let's consider the interval $[0, 2\pi]$ and then use symmetry. However, the integrand is defined on $[0, \pi]$.

Let's go back to the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$. $I = \int_0^\pi \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} dx$. Let $f(x) = \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x}$. $f(\pi - x) = \frac{(\pi - x)^2 \sin(\pi - x) \cos(\pi - x)}{\sin^4(\pi - x) + \cos^4(\pi - x)} = \frac{(\pi - x)^2 (-\sin x \cos x)}{\sin^4 x + \cos^4 x}$. $I = \int_0^\pi f(\pi - x) dx = \int_0^\pi \frac{(\pi - x)^2 (-\sin x \cos x)}{\sin^4 x + \cos^4 x} dx$. $2I = \int_0^\pi \frac{x^2 \sin x \cos x - (\pi - x)^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx$ $2I = \int_0^\pi \frac{(x^2 - (\pi - x)^2) \sin x \cos x}{\sin^4 x + \cos^4 x} dx$ $2I = \int_0^\pi \frac{(2\pi x - \pi^2) \sin x \cos x}{\sin^4 x + \cos^4 x} dx$ $2I = \int_0^\pi \frac{2\pi x \sin x \cos x}{\sin^4 x + \cos^4 x} dx - \pi^2 \int_0^\pi \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx$.

Let $J = \int_0^\pi \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx$. Let's consider the integrand $g(x) = \frac{\sin x \cos x}{\sin^4 x + \cos^4 x}$. $g(\pi - x) = \frac{\sin(\pi - x) \cos(\pi - x)}{\sin^4(\pi - x) + \cos^4(\pi - x)} = \frac{(\sin x)(-\cos x)}{\sin^4 x + \cos^4 x} = -g(x)$. For an integral $\int_0^{2a} f(x) dx$, if $f(2a-x) = -f(x)$, then the integral is 0. Here, the integral is from $0$ to $\pi$. Let's consider $2a = \pi$, so $a = \pi/2$. Let's consider the integral from $0$ to $\pi$. The property is: if $f(\pi - x) = -f(x)$, then $\int_0^\pi f(x) dx = 0$. Here, $g(\pi - x) = -g(x)$, so $J = \int_0^\pi \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx = 0$.

Now consider the first part of $2I$: $\int_0^\pi \frac{2\pi x \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. Let $k(x) = \frac{2\pi x \sin x \cos x}{\sin^4 x + \cos^4 x}$. $k(\pi - x) = \frac{2\pi (\pi - x) \sin(\pi - x) \cos(\pi - x)}{\sin^4(\pi - x) + \cos^4(\pi - x)} = \frac{2\pi (\pi - x) (-\sin x \cos x)}{\sin^4 x + \cos^4 x}$. Let $I' = \int_0^\pi \frac{2\pi x \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. $I' = \int_0^\pi k(\pi - x) dx = \int_0^\pi \frac{2\pi (\pi - x) (-\sin x \cos x)}{\sin^4 x + \cos^4 x} dx$. $2I' = \int_0^\pi \frac{2\pi x \sin x \cos x}{\sin^4 x + \cos^4 x} dx + \int_0^\pi \frac{-2\pi (\pi - x) \sin x \cos x}{\sin^4 x + \cos^4 x} dx$ $2I' = \int_0^\pi \frac{[2\pi x - 2\pi (\pi - x)] \sin x \cos x}{\sin^4 x + \cos^4 x} dx$ $2I' = \int_0^\pi \frac{(2\pi x - 2\pi^2 + 2\pi x) \sin x \cos x}{\sin^4 x + \cos^4 x} dx$ $2I' = \int_0^\pi \frac{(4\pi x - 2\pi^2) \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. This is still not simplifying to zero easily.

Let's use the property $\int_0^{2a} f(x) dx = a \int_0^2 f(x) dx$ if $f(2a-x)=f(x)$. And $\int_0^{2a} f(x) dx = 0$ if $f(2a-x)=-f(x)$.

Let's go back to $2I = \int_0^\pi \frac{(2\pi x - \pi^2) \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. Let $h(x) = \frac{(2\pi x - \pi^2) \sin x \cos x}{\sin^4 x + \cos^4 x}$. We found that $h(\pi - x) = \frac{(2\pi (\pi - x) - \pi^2) (-\sin x \cos x)}{\sin^4 x + \cos^4 x} = \frac{(2\pi^2 - 2\pi x - \pi^2) (-\sin x \cos x)}{\sin^4 x + \cos^4 x} = \frac{(\pi^2 - 2\pi x) (-\sin x \cos x)}{\sin^4 x + \cos^4 x} = \frac{-(2\pi x - \pi^2) (-\sin x \cos x)}{\sin^4 x + \cos^4 x} = \frac{(2\pi x - \pi^2) \sin x \cos x}{\sin^4 x + \cos^4 x} = h(x)$.

So, for the integral $\int_0^\pi h(x) dx$, we have $h(\pi - x) = h(x)$. This means $\int_0^\pi h(x) dx = 2 \int_0^{\pi/2} h(x) dx$. This does not directly lead to zero.

Let's reconsider the initial integral I = \int_\limits0^\pi \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x. Let $u = \sin^2 x$. Then $du = 2 \sin x \cos x dx$. When $x=0$, $u=0$. When $x=\pi$, $u=0$. This substitution won't work directly because the limits become the same.

Let's use the property $\int_0^{2a} f(x) dx = \int_0^{2a} f(2a-x) dx$. Let's consider the integrand $F(x) = \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x}$. We need to evaluate $\left|\frac{120}{\pi^3} I\right|$.

Let's consider the integral $\int_0^\pi x \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx$. Let $g(x) = \frac{\sin x \cos x}{\sin^4 x + \cos^4 x}$. We know $g(\pi - x) = -g(x)$. So, $\int_0^\pi x g(x) dx$. Let $I_x = \int_0^\pi x g(x) dx$. $I_x = \int_0^\pi (\pi - x) g(\pi - x) dx = \int_0^\pi (\pi - x) (-g(x)) dx = -\int_0^\pi (\pi - x) g(x) dx$. $I_x = -\pi \int_0^\pi g(x) dx + \int_0^\pi x g(x) dx$. Since $\int_0^\pi g(x) dx = 0$, we have $I_x = I_x$. This doesn't help.

Let's go back to $2I = \int_0^\pi \frac{(2\pi x - \pi^2) \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. Let $u = \tan x$. Then $du = \sec^2 x dx$. This is not suitable.

Consider the substitution $t = \tan x$. $\sin^2 x = \frac{t^2}{1+t^2}$, $\cos^2 x = \frac{1}{1+t^2}$. $\sin^4 x + \cos^4 x = \left(\frac{t^2}{1+t^2}\right)^2 + \left(\frac{1}{1+t^2}\right)^2 = \frac{t^4 + 1}{(1+t^2)^2}$. $\sin x \cos x = \frac{t}{1+t^2}$. When $x$ goes from $0$ to $\pi$, $t$ goes from $0$ to $\infty$ and then from $-\infty$ to $0$. This is problematic.

Let's split the integral: $I = \int_0^{\pi/2} \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x + \int_{\pi/2}^\pi \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x$. In the second integral, let $x = \pi - y$. Then $dx = -dy$. When $x=\pi/2$, $y=\pi/2$. When $x=\pi$, $y=0$. $\int_{\pi/2}^\pi \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x = \int_{\pi/2}^0 \frac{(\pi - y)^2 \sin(\pi - y) \cos(\pi - y)}{\sin^4(\pi - y) + \cos^4(\pi - y)} (-dy)$ $= \int_0^{\pi/2} \frac{(\pi - y)^2 (-\sin y \cos y)}{\sin^4 y + \cos^4 y} dy = -\int_0^{\pi/2} \frac{(\pi - y)^2 \sin y \cos y}{\sin^4 y + \cos^4 y} dy$. So, $I = \int_0^{\pi/2} \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x - \int_0^{\pi/2} \frac{(\pi - x)^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x$. $I = \int_0^{\pi/2} \frac{[x^2 - (\pi - x)^2] \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. $I = \int_0^{\pi/2} \frac{(2\pi x - \pi^2) \sin x \cos x}{\sin^4 x + \cos^4 x} dx$.

Let $u = \tan x$. Then $du = \sec^2 x dx$. $\sin x \cos x = \frac{\tan x}{\sec^2 x} = \frac{u}{1+u^2}$. $\sin^4 x + \cos^4 x = \frac{u^4 + 1}{(1+u^2)^2}$. $\sin x \cos x dx = \frac{u}{1+u^2} \frac{du}{\sec^2 x} = \frac{u}{1+u^2} \frac{du}{1+u^2} = \frac{u}{(1+u^2)^2} du$. When $x=0$, $u=0$. When $x=\pi/2$, $u \to \infty$. $I = \int_0^\infty \frac{(2\pi \arctan u - \pi^2) \frac{u}{(1+u^2)^2}}{\frac{u^4 + 1}{(1+u^2)^2}} du$ $I = \int_0^\infty (2\pi \arctan u - \pi^2) \frac{u}{u^4 + 1} du$. $I = 2\pi \int_0^\infty \frac{u \arctan u}{u^4 + 1} du - \pi^2 \int_0^\infty \frac{u}{u^4 + 1} du$.

Consider the integral $\int_0^\infty \frac{u}{u^4 + 1} du$. Let $v = u^2$, $dv = 2u du$. $\int_0^\infty \frac{1}{v^2+1} \frac{dv}{2} = \frac{1}{2} [\arctan v]_0^\infty = \frac{1}{2} (\frac{\pi}{2} - 0) = \frac{\pi}{4}$. So, $\pi^2 \int_0^\infty \frac{u}{u^4 + 1} du = \pi^2 \frac{\pi}{4} = \frac{\pi^3}{4}$.

Now consider $2\pi \int_0^\infty \frac{u \arctan u}{u^4 + 1} du$. Let's examine the original integral. If we consider the integrand $F(x) = \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x}$. Let's check the property $F(\pi - x) = -F(x)$ for the entire interval $[0, \pi]$. $F(\pi - x) = \frac{(\pi - x)^2 \sin(\pi - x) \cos(\pi - x)}{\sin^4(\pi - x) + \cos^4(\pi - x)} = \frac{(\pi - x)^2 (-\sin x \cos x)}{\sin^4 x + \cos^4 x}$. This is not $-F(x)$.

Let's consider the integral $I = \int_0^\pi \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. Let's observe the symmetry around $x = \pi/2$. Let $x = \pi/2 + t$. Then $dx = dt$. When $x=0$, $t=-\pi/2$. When $x=\pi$, $t=\pi/2$. $I = \int_{-\pi/2}^{\pi/2} \frac{(\pi/2 + t)^2 \sin(\pi/2 + t) \cos(\pi/2 + t)}{\sin^4(\pi/2 + t) + \cos^4(\pi/2 + t)} dt$. $\sin(\pi/2 + t) = \cos t$. $\cos(\pi/2 + t) = -\sin t$. $\sin^4(\pi/2 + t) = \cos^4 t$. $\cos^4(\pi/2 + t) = \sin^4 t$. So, the integral becomes: $I = \int_{-\pi/2}^{\pi/2} \frac{(\pi/2 + t)^2 (\cos t)(-\sin t)}{\cos^4 t + \sin^4 t} dt$. $I = -\int_{-\pi/2}^{\pi/2} \frac{(\pi/2 + t)^2 \sin t \cos t}{\cos^4 t + \sin^4 t} dt$. Let $g(t) = \frac{\sin t \cos t}{\cos^4 t + \sin^4 t}$. This is an odd function of $t$ because $\sin(-t) \cos(-t) = (-\sin t)(\cos t) = -\sin t \cos t$, and $\cos^4(-t) + \sin^4(-t) = \cos^4 t + \sin^4 t$. So, $\int_{-\pi/2}^{\pi/2} g(t) dt = 0$.

Let's expand $(\pi/2 + t)^2 = (\pi/2)^2 + \pi t + t^2$. $I = -\int_{-\pi/2}^{\pi/2} \frac{[(\pi/2)^2 + \pi t + t^2] \sin t \cos t}{\cos^4 t + \sin^4 t} dt$. $I = -\int_{-\pi/2}^{\pi/2} \frac{(\pi/2)^2 \sin t \cos t}{\cos^4 t + \sin^4 t} dt - \int_{-\pi/2}^{\pi/2} \frac{\pi t \sin t \cos t}{\cos^4 t + \sin^4 t} dt - \int_{-\pi/2}^{\pi/2} \frac{t^2 \sin t \cos t}{\cos^4 t + \sin^4 t} dt$.

The second integral $\int_{-\pi/2}^{\pi/2} \frac{\pi t \sin t \cos t}{\cos^4 t + \sin^4 t} dt = 0$ because the integrand is an odd function of $t$. The first integral $\int_{-\pi/2}^{\pi/2} \frac{(\pi/2)^2 \sin t \cos t}{\cos^4 t + \sin^4 t} dt$. Let $G(t) = \frac{\sin t \cos t}{\cos^4 t + \sin^4 t}$. $G(-t) = -G(t)$, so $G(t)$ is odd. Thus, $\int_{-\pi/2}^{\pi/2} G(t) dt = 0$. Therefore, the first integral is 0.

Now consider the third integral: $-\int_{-\pi/2}^{\pi/2} \frac{t^2 \sin t \cos t}{\cos^4 t + \sin^4 t} dt$. Let $H(t) = \frac{t^2 \sin t \cos t}{\cos^4 t + \sin^4 t}$. $H(-t) = \frac{(-t)^2 \sin(-t) \cos(-t)}{\cos^4(-t) + \sin^4(-t)} = \frac{t^2 (-\sin t) (\cos t)}{\cos^4 t + \sin^4 t} = -\frac{t^2 \sin t \cos t}{\cos^4 t + \sin^4 t} = -H(t)$. So, the third integral is also 0.

Thus, $I = 0$. The expression we need to evaluate is $\left|\frac{120}{\pi^3} I\right|$. Since $I=0$, the value is $\left|\frac{120}{\pi^3} \times 0\right| = 0$.

Common Mistakes & Tips

• Be careful when applying the King's property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ and the property $\int_0^{2a} f(x) dx = 0$ if $f(2a-x) = -f(x)$. Ensure the conditions are met for the correct integral limits.
• Symmetry arguments are powerful. Splitting the integral interval or using substitutions like $x = a+t$ can reveal useful symmetries.
• When dealing with trigonometric integrands, simplifying the denominator using identities like $\sin^4 x + \cos^4 x = 1 - \frac{1}{2} \sin^2(2x)$ can be helpful.

Summary

The problem involves evaluating a definite integral with trigonometric functions. We used the property of definite integrals related to symmetry around the midpoint of the integration interval. By substituting $x = \pi/2 + t$, we transformed the integral from $[0, \pi]$ to $[-\pi/2, \pi/2]$. The resulting integrand was analyzed for its symmetry properties. We found that the integral could be broken down into components, each of which evaluated to zero due to the integrand being an odd function over a symmetric interval. Consequently, the value of the integral $I$ is 0, making the final expression also 0.

The final answer is $\boxed{0}$.