Answer
$-\frac{1}{5}\cos^5 t+\frac{2}{7}\cos^7 t-\frac{1}{9}\cos^9 t+C$
Work Step by Step
$\int\sin^5 t\cos^4 t\ dt$
Since the power of sine is odd, save a factor of $\sin t$ and express the remaining factors in terms of $\cos t$:
$=\int\sin^4 t\cos^4 t\sin t\ dt$
$=\int(\sin^2 t)^2\cos^4 t\sin t\ dt$
$=\int(1-\cos^2 t)^2\cos^4 t\sin t\ dt$
Let $u=\cos t$. Then $du=-\sin t\ dt$, and $\sin t\ dt=-du$.
$=\int(1-u^2)^2 u^4*(-1)\ du$
$=-\int(1-2u^2+u^4)u^4\ du$
$=-\int(u^4-2u^6+u^8)\ du$
$=-(\frac{1}{5}u^5-\frac{2}{7}u^7+\frac{1}{9}u^9+C)$
$=-(\frac{1}{5}\cos^5 t-\frac{2}{7}\cos^7 t+\frac{1}{9}\cos^9 t+C)$
$=\boxed{-\frac{1}{5}\cos^5 t+\frac{2}{7}\cos^7 t-\frac{1}{9}\cos^9 t+C}$
