Answer
$\frac{1}{7}\cos^7\theta-\frac{1}{5}\cos^5\theta+C$
Work Step by Step
$\int \sin^3\theta\cos^4\theta\ d\theta$
Since the power of sine is odd, save a factor of $\sin\theta$ and express the rest in terms of $\cos \theta$:
$=\int\sin^2\theta\sin\theta\cos^4\theta\ d\theta$
$=\int(1-\cos^2\theta)\cos^4\theta\sin\theta\ d\theta$
Let $u=\cos\theta$. Then $du=-\sin\theta\ d\theta$, and $\sin\theta\ d\theta=-du$
$=\int(1-u^2)u^4*(-1)du$
$=\int(u^4-u^6)*(-1)du$
$=\int(u^6-u^4)du$
$=\frac{1}{7}u^7-\frac{1}{5}u^5+C$
$=\boxed{\frac{1}{7}\cos^7\theta-\frac{1}{5}\cos^5\theta+C}$
