Answer
a) $\int$sin(x)cos(x)dx, Solve using u-substitution; u=sin(x)
=$-\frac{sin^2(x)}{2}+c$
b) $\int$sin(x)cos(x)dx, Solve using the trig identity; sin(2x)=2sin(x)cos(x)
=$-\frac{cos(2x)}{4}+c$
Work Step by Step
a) $\int$sin(x)cos(x)dx, Solve using u-substitution; u=sin(x)
=$\int ucos(x)dx$ u=sin(x), -du=cos(x)dx
=$-\int udu$
=$-\frac{u^2}{2}+c$
=$-\frac{sin^2(x)}{2}+c$
b) $\int$sin(x)cos(x)dx, Solve using the trig identity; sin(2x)=2sin(x)cos(x)
= $\frac{1}{2}\int sin(2x)$, (trig identity: sin(2x)=2sin(x)cos(x) => $\frac{sin(2x)}{2}=sin(x)cos(x)$)
=$\frac{1}{2}sin(u)dx$, u=2x, $\frac{du}{2}=dx$
=$\frac{1}{4}\int sin(u)du$
=$-\frac{1}{4}cos(u)+c$
=$-\frac{cos(2x)}{4}+c$
