Question 54

Given the definite integral:

1. Substitution:
   Let $u = \frac{1}{x}$. Then:

   $$du = -\frac{1}{x^2}\,dx \implies \frac{1}{x^2}\,dx = -du.$$

2. Change of integration limits:

   • Lower limit: when $x = 1/\pi$, $u = \pi$.
   • Upper limit: when $x = 2/\pi$, $u = \pi/2$.

3. Evaluation:

   $$I = \int_{\pi}^{\pi/2} \cos(u)\,(-du) = \int_{\pi/2}^{\pi} \cos(u)\,du$$
   $$I = \left[ \sin(u) \right]_{\pi/2}^{\pi} = \sin(\pi) - \sin(\pi/2) = 0 - 1 = -1.$$

The value of the integral is $-1$.