Integral

First let us substitute $2^{x} = t$

So the integral becomes:-

$\large \frac{1}{\sin(1)\ln(2)} \int_{0}^{\infty} \frac{\sin^{2}(t)}{t^{2}} dt$

$\large = \frac{1}{2\sin(1)\ln(2)} \int_{0}^{\infty} \frac{1-\cos(2t)}{t^{2}} dt$

Let $\large I(a) = \frac{1}{2\sin(1)\ln(2)} \int_{0}^{\infty} \frac{1-\cos(at)}{t^{2}} dt$

So $\large I'(a) = \frac{1}{2\sin(1)\ln(2)} \int_{0}^{\infty} \frac{t\sin(at)}{t^{2}} dt$

$\large I'(a) = \frac{1}{2\sin(1)\ln(2)} \int_{0}^{\infty} \frac{\sin(at)}{t} dt$

again using a substitution of $at = z$

We have:- $\large I'(a) = \frac{1}{2\sin(1)\ln(2)} \int_{0}^{\infty} \frac{\sin(z)}{z} dz$

We know $\large \int_{0}^{\infty} \frac{\sin(x)}{x} dx$ is the Dirichlet Integral and it's value is $\frac{\pi}{2}$

So we have $\large I'(a) = \frac{1}{2\sin(1)\ln(2)}\frac{\pi}{2}$

So $\large I(a) = \frac{1}{2\sin(1)\ln(2)}(\frac{a\pi}{2} + C)$

We have $I(0) = 0$

So $C=0$

So we have $\large I(a) = \frac{1}{2\sin(1)\ln(2)}(\frac{a\pi}{2})$

So $\large I(2) = \frac{1}{\sin(1)\ln(2)}(\frac{\pi}{2})$

Which is our required answer.