Property: f(a+b-x)

Let

$$I=\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4x+\cos^4x}\,dx$$

Using

$$\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx,$$

we get

$$I= \int_0^{\pi/2} \frac{\left(\frac{\pi}{2}-x\right)\cos x\sin x}{\cos^4x+\sin^4x}\,dx$$

Adding the two forms,

$$2I= \frac{\pi}{2}\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}\,dx$$

Divide the numerator and denominator by $$\cos^4x$$:

$$2I= \frac{\pi}{2}\int_0^{\pi/2}\frac{\tan x\sec^2x}{\tan^4x+1}\,dx$$

Put

$$u=\tan^2x,$$

then

$$du=2\tan x\sec^2x\,dx \implies \tan x\sec^2x\,dx = \frac{du}{2}$$

Changing the limits of integration:
When $$x = 0 \implies u = 0$$
When $$x = \frac{\pi}{2} \implies u \to \infty$$

Therefore,

$$2I= \frac{\pi}{2}\int_0^\infty\frac{1}{1+u^2}\cdot\frac{du}{2}$$

$$2I= \frac{\pi}{4}\left[\tan^{-1}u\right]_0^\infty$$

$$2I= \frac{\pi}{4}\cdot\frac{\pi}{2}$$

$$2I= \frac{\pi^2}{8}$$

Hence,

$$I=\frac{\pi^2}{16}$$

Since $$\lambda = I = \frac{\pi^2}{16}$$, we have:

$$\frac{32\lambda}{\pi^2}=2$$

Let

$$I = \int_0^1 \frac{\ln(1+x)}{1+x^2}\,dx$$

Substitute $$x = \tan\theta$$, which gives $$dx = \sec^2\theta\,d\theta$$.

Changing the limits of integration:
When $$x = 0 \implies \theta = 0$$
When $$x = 1 \implies \theta = \frac{\pi}{4}$$

Substituting these into the integral:

$$I = \int_0^{\pi/4} \frac{\ln(1+\tan\theta)}{1+\tan^2\theta} \cdot \sec^2\theta\,d\theta$$

Since $$1+\tan^2\theta = \sec^2\theta$$, the terms cancel out:

$$I = \int_0^{\pi/4} \ln(1+\tan\theta)\,d\theta \quad \cdots(1)$$

Using the property $$\int_0^a f(\theta)\,d\theta = \int_0^a f(a-\theta)\,d\theta$$ with $$\theta \mapsto \frac{\pi}{4}-\theta$$, we get:

$$I = \int_0^{\pi/4} \ln\left(1+\tan\left(\frac{\pi}{4}-\theta\right)\right)\,d\theta$$

Using the trigonometric identity $$\tan\left(\frac{\pi}{4}-\theta\right) = \frac{1-\tan\theta}{1+\tan\theta}$$:

$$I = \int_0^{\pi/4} \ln\left(1 + \frac{1-\tan\theta}{1+\tan\theta}\right)\,d\theta$$

$$I = \int_0^{\pi/4} \ln\left(\frac{1+\tan\theta + 1-\tan\theta}{1+\tan\theta}\right)\,d\theta$$

$$I = \int_0^{\pi/4} \ln\left(\frac{2}{1+\tan\theta}\right)\,d\theta$$

Using the logarithmic property $$\ln\left(\frac{a}{b}\right) = \ln a - \ln b$$:

$$I = \int_0^{\pi/4} \left[\ln 2 - \ln(1+\tan\theta)\right]\,d\theta \quad \cdots(2)$$

Adding equation (1) and equation (2):

$$2I = \int_0^{\pi/4} \ln 2\,d\theta$$

$$2I = \ln 2 \cdot [\theta]_0^{\pi/4}$$

$$2I = \frac{\pi}{4}\ln 2$$

Hence,

$$I = \frac{\pi}{8}\ln 2$$

Since $$\lambda = I = \frac{\pi}{8}\ln 2$$, we have:

$$\frac{8\lambda}{\pi\ \ln2}=\frac{8}{\pi\ \ln2}\left(\frac{\pi}{8}\ln2\right)=1$$

We are given the functional differential equation:

$$f(x) \cdot f'(-x) = f(-x) \cdot f'(x)$$

Rearranging all terms to one side:

$$f(-x) \cdot f'(x) - f(x) \cdot f'(-x) = 0$$

Notice that this expression is the exact derivative of the product $$f(x) \cdot f(-x)$$ using the chain rule:

$$\frac{d}{dx} [f(x) \cdot f(-x)] = f'(x) \cdot f(-x) + f(x) \cdot \left(f'(-x) \cdot (-1)\right) = f(-x) \cdot f'(x) - f(x) \cdot f'(-x)$$

Therefore, we can rewrite the equation as:

$$\frac{d}{dx} [f(x) \cdot f(-x)] = 0$$

Integrating both sides with respect to $$x$$:

$$f(x) \cdot f(-x) = C \quad \text{(where } C \text{ is a constant)}$$

To find the value of $$C$$, we use the initial condition $$f(0) = 3$$:

$$f(0) \cdot f(-0) = C$$ $$3 \cdot 3 = C \implies C = 9$$

Thus, we have the fundamental relationship:

$$f(x) \cdot f(-x) = 9$$

Let the given integral be $$I$$:

$$I = \int_{-51}^{51} \frac{1}{3 + f(x)} \, dx \quad \text{--- (Equation 1)}$$

Using King's Property of definite integrals, $$\int_{a}^{b} g(x) \, dx = \int_{a}^{b} g(a + b - x) \, dx$$, we substitute $$x$$ with $$(-51 + 51 - x) = -x$$:

$$I = \int_{-51}^{51} \frac{1}{3 + f(-x)} \, dx \quad \text{--- (Equation 2)}$$

From our finding in Step 1, we know that $$f(-x) = \frac{9}{f(x)}$$. Substituting this into Equation 2:

$$I = \int_{-51}^{51} \frac{1}{3 + \frac{9}{f(x)}} \, dx$$

Taking the common denominator in the integrand:

$$I = \int_{-51}^{51} \frac{f(x)}{3f(x) + 9} \, dx = \int_{-51}^{51} \frac{f(x)}{3[3 + f(x)]} \, dx \quad \text{--- (Equation 3)}$$

Now, add Equation 1 and Equation 3 together:

$$I + I = \int_{-51}^{51} \frac{1}{3 + f(x)} \, dx + \int_{-51}^{51} \frac{f(x)}{3[3 + f(x)]} \, dx$$

Combine the terms under a single integral by making a common denominator ($$3[3 + f(x)]$$):

$$2I = \int_{-51}^{51} \frac{3 + f(x)}{3[3 + f(x)]} \, dx$$

The terms $$(3 + f(x))$$ cancel out perfectly from both the numerator and denominator:

$$2I = \int_{-51}^{51} \frac{1}{3} \, dx$$ $$2I = \frac{1}{3} \Big[ x \Big]_{-51}^{51}$$ $$2I = \frac{1}{3} \Big[ 51 - (-51) \Big]$$ $$2I = \frac{1}{3} \cdot 102$$ $$2I = 34$$ 

$$I = 17$$