[tex]\bf \displaystyle\int sin(kx)\implies \cfrac{-cos(kx)}{k} + C \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \displaystyle\int\limits_{0}^{5}sin(3x)dx\implies \left. \cfrac{-cos(3x)}{3} \right]_{0}^{5}\implies \left[ \cfrac{-cos(15)}{3} \right]-\left[ -\cfrac{1}{3} \right]\implies \cfrac{1-cos(15)}{3}[/tex]
