Integrating $\int\sin3\theta \cos\theta \, d\theta$ using factor formulae vs trigonometric identity expansion?

Question

When you perform $\int\sin3\theta \cos\theta \, d\theta$ using factor formulae you easily get $-\frac{1}{8} \cos4\theta-\frac{1}{4}\cos2\theta + K$. However, when I used expansion using trig identities I got: \begin{align} \int\sin3\theta \cos\theta \, d\theta &= \int \sin(2\theta +\theta) \cos\theta \, d\theta \\ &= \int(\sin2\theta \cos\theta + \cos2\theta \sin \theta)\cos\theta \, d\theta \\ &= \int[2\sin\theta \cos\theta \cos\theta + (1-2\sin^2\theta) \sin\theta]\cos\theta \, d\theta \\ &= \int[2\sin\theta \cos^2\theta + \sin\theta -2\sin^3\theta]\cos\theta \, d\theta \\ &= \int (2\sin\theta \cos^3\theta + \sin\theta\cos\theta -2\cos\theta\sin^3\theta) \, d\theta \\ &= 2 \, \left(-\frac{1}{4}\cos^4\theta \right) + \frac{1}{2} \sin^2\theta -2 \, \left(\frac{1}{4} \sin^4\theta \right) +K \\ &= -\frac{1}{2}\cos^4\theta + \frac{1}{2} \sin^2\theta -\frac{1}{2} \sin^4\theta +K \end{align} But after substituting with particular values for $\theta$ they don't give the same answer, so why don't they agree?