Double Integral Evaluation - Solution to Exercise 6

$$I = \int_{s=0}^{\pi} \int_{t=0}^{s} s \cos(\pi - t) \, dt \, ds$$

$$= \int_{0}^{\pi} \Bigg[ -s \sin(\pi - t) \Bigg]_{t=0}^{t=s} \, ds = -\int_{0}^{\pi} s \Bigg[ \sin(\pi - s) - \sin(\pi) \Bigg] \, ds = -\int_{0}^{\pi} s \sin(\pi - s) \, ds$$

$$\text{Since } \sin(\pi - s) = \sin(s), \text{ we have: } I = -\int_{0}^{\pi} s \sin(s) \, ds$$

$$\text{Integrate by parts: Let } u = s, \; dv = \sin(s) ds \Rightarrow du = ds, \; v = -\cos(s)$$

$$I = -\Bigg[ -s \cos(s) \Bigg]_{0}^{\pi} - \int_{0}^{\pi} (-\cos(s)) \, ds = -\Bigg[ -\pi \cos(\pi) - 0 \Bigg] - \Bigg[ \sin(s) \Bigg]_{0}^{\pi} = -\Bigg( \pi - 0 \Bigg) - (0 - 0) = -\pi$$