Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 2

Determine the integral $\displaystyle \int \sin^6 x \cos ^3 x dx$


$
\begin{equation}
\begin{aligned}

\int \sin^6 x \cos ^3 x dx =& \int \sin^6 x \cos^2 x \cos x dx
\qquad \text{Apply Trigonometric Identity } \cos^2 x = 1 - \sin^2 x
\\
\\
\int \sin^6 x \cos ^3 x dx =& \int \sin^6 x (1 - \sin^2 x) \cos x dx
\\
\\
\int \sin^6 x \cos ^3 x dx =& \int (\sin^6 x - \sin^8 x) \cos x dx


\end{aligned}
\end{equation}
$


Let $u = \sin x$, then $du = \cos x dx$. Thus,


$
\begin{equation}
\begin{aligned}

\int \sin^6 x \cos ^3 x dx =& \int u^6 - u^8 du
\\
\\
\int \sin^6 x \cos ^3 x dx =& \frac{u^{6 + 1}}{6 + 1} - \frac{u^{8 + 1}}{8 + 1} + c
\\
\\
\int \sin^6 x \cos ^3 x dx =& \frac{u^7}{7} - \frac{u^9}{9} + c
\\
\\
\int \sin^6 x \cos ^3 x dx =& \frac{(\sin x)^7}{7} - \frac{(\sin x)^9}{9} + c
\\
\\
\int \sin^6 x \cos ^3 x dx =& \frac{\sin ^7 x}{7} - \frac{\sin^9 x}{9} + c

\end{aligned}
\end{equation}
$