Double Integral Evaluation - Solution to Exercise 4

$$I = \int_{0}^{\pi} \sin(2x) \Bigg( \int_{0}^{x} dy \Bigg) dx = \int_{0}^{\pi} x \sin(2x) \, dx \quad \text{(Integrate by parts)}$$

$$\text{Let } u = x, \; dv = \sin(2x) dx \; \Rightarrow \; du = dx, \; v = -\frac{1}{2}\cos(2x)$$

$$I = \Bigg[ -\frac{x}{2}\cos(2x) \Bigg]_{0}^{\pi} - \int_{0}^{\pi} \Bigg( -\frac{1}{2}\cos(2x) \Bigg) dx = \Bigg( -\frac{\pi}{2}\cos(2\pi) - 0 \Bigg) - \Bigg( -\frac{1}{2}\sin(2x) \Bigg)_{0}^{\pi}$$

$$= -\frac{\pi}{2} - 0 = -\frac{\pi}{2}$$