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Step-by-Step Solution
Step 1: Analyze the function f(x) given its integrand
We have
f(x) = \int_{0}^{x} g(t) \,\ln\!\biggl(\frac{1 - t}{1 + t}\biggr) \, dt,
where g(t) is a continuous odd function (that is, g(-t) = -g(t) ).
Next, note that
\ln\!\biggl(\frac{1 - t}{1 + t}\biggr)
is also an odd function of t . Specifically,
\ln\!\biggl(\frac{1 - (-t)}{1 + (-t)}\biggr)
= \ln\!\biggl(\frac{1 + t}{1 - t}\biggr)
= -\,\ln\!\biggl(\frac{1 - t}{1 + t}\biggr).
Hence, the product
g(t)\,\ln\!\biggl(\frac{1 - t}{1 + t}\biggr)
is even (odd × odd = even). Integrating an even function from 0 to x gives an odd function f(x) . Thus f(-x) = -f(x) , and for any symmetric interval about 0:
\int_{-a}^{a} f(x)\,dx = 0.
Step 2: Express the required definite integral
We need to evaluate
\int_{-\pi/2}^{\pi/2} \Bigl[f(x) + \frac{x^2 \cos x}{1 + e^x}\Bigr] \, dx.
Since f(x) is odd,
\int_{-\pi/2}^{\pi/2} f(x)\, dx = 0.
Therefore, the given integral simplifies to
\int_{-\pi/2}^{\pi/2} \frac{x^2 \cos x}{1 + e^x}\, dx.
Step 3: Simplify the integral using symmetry properties
Let
I = \int_{-\pi/2}^{\pi/2} \frac{x^2 \cos x}{1 + e^x}\, dx.
Split this into two parts:
I = \int_{-\pi/2}^{0} \frac{x^2 \cos x}{1 + e^x}\, dx
\;+\; \int_{0}^{\pi/2} \frac{x^2 \cos x}{1 + e^x}\, dx.
In the first integral, let x = -u \implies dx = -du . When x = -\frac{\pi}{2} , u = \frac{\pi}{2} ; when x = 0 , u = 0 . Thus:
\int_{-\pi/2}^{0} \frac{x^2 \cos x}{1 + e^x}\, dx
= \int_{\pi/2}^{0} \frac{(-u)^2 \cos(-u)}{1 + e^{-u}} \, (-du).
Since (-u)^2 = u^2 and \cos(-u) = \cos(u) , and also
1 + e^{-u} = \frac{1 + e^u}{e^u} \implies \frac{1}{1 + e^{-u}} = \frac{e^u}{1 + e^u},
we get
\int_{\pi/2}^{0} \frac{u^2 \cos u \, e^u}{1 + e^u} (-du)
= \int_{0}^{\pi/2} \frac{u^2 \cos u \, e^u}{1 + e^u}\, du.
Hence,
I = \int_{0}^{\pi/2} \frac{x^2 \cos x}{1 + e^x} \, dx
\;+\; \int_{0}^{\pi/2} \frac{x^2 \cos x \, e^x}{1 + e^x} \, dx.
Combine the integrands:
I = \int_{0}^{\pi/2} x^2 \cos x \,\Bigl[\frac{1}{1 + e^x} + \frac{e^x}{1 + e^x}\Bigr] \, dx
= \int_{0}^{\pi/2} x^2 \cos x \,\Bigl[\frac{1 + e^x}{1 + e^x}\Bigr] \, dx
= \int_{0}^{\pi/2} x^2 \cos x \, dx.
Step 4: Evaluate \int_{0}^{\pi/2} x^2 \cos x \, dx
We use integration by parts twice. Let us set up the first integration by parts:
u = x^2 \implies du = 2x \, dx.
dv = \cos x \, dx \implies v = \sin x.
Then
\int x^2 \cos x \, dx
= x^2 \sin x - \int 2x \sin x \, dx.
For the remaining integral \int 2x \sin x \, dx , factor out 2:
\int 2x \sin x \, dx = 2 \int x \sin x \, dx.
Now apply integration by parts again to \int x \sin x \, dx :
u = x \implies du = dx.
dv = \sin x \, dx \implies v = -\cos x.
So
\int x \sin x \, dx
= x \cdot (-\cos x) - \int 1 \cdot (-\cos x)\, dx
= -\,x \cos x + \int \cos x \, dx
= -\,x \cos x + \sin x.
Therefore,
\int 2x \sin x \, dx
= 2 \bigl[-x \cos x + \sin x\bigr]
= -\,2x \cos x + 2 \sin x.
Substitute back:
\int x^2 \cos x \, dx
= x^2 \sin x - \bigl[-2x \cos x + 2 \sin x\bigr]
= x^2 \sin x + 2x \cos x - 2 \sin x + C.
Now evaluate from 0 to \frac{\pi}{2} :
\int_{0}^{\pi/2} x^2 \cos x \, dx
= \bigl[x^2 \sin x + 2x \cos x - 2 \sin x\bigr]_{0}^{\pi/2}.
At x = \frac{\pi}{2} :
\left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) = \frac{\pi^2}{4}\cdot 1 = \frac{\pi^2}{4}.
2 \cdot \frac{\pi}{2} \cos\left(\frac{\pi}{2}\right) = \pi \cdot 0 = 0.
-2 \sin\left(\frac{\pi}{2}\right) = -2 \cdot 1 = -2.
Hence the expression at x = \frac{\pi}{2} is \frac{\pi^2}{4} - 2.
At x = 0 :
Everything is zero.
Thus,
\int_{0}^{\pi/2} x^2 \cos x \, dx = \frac{\pi^2}{4} - 2.
Step 5: Solve for \alpha using the equality given in the problem
So the value of the original integral is
\frac{\pi^2}{4} - 2.
According to the question:
\int_{-\pi/2}^{\pi/2}\Bigl(f(x)+\frac{x^2 \cos x}{1+ e^x}\Bigr) dx
= \Bigl(\frac{\pi}{\alpha}\Bigr)^2 - \alpha
= \frac{\pi^2}{4} - 2.
Rewrite the left side as
\frac{\pi^2}{\alpha^2},
so
\frac{\pi^2}{\alpha^2} - \alpha = \frac{\pi^2}{4} - 2.
By inspection or direct solving, \alpha = 2 satisfies this equation. Indeed, if \alpha = 2 :
Left-hand side = \frac{\pi^2}{4} - 2 , which matches the right-hand side.
Therefore,
\alpha = 2.
Final Answer
The value of \alpha is 2.
