Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 52

Find the indefinite integral $\displaystyle \int \sin^3 x \cos^4 x dx$. Illustrate by graphing both the integrand and its antiderivative (taking $c = 0$).

$
\begin{equation}
\begin{aligned}
\int \sin^3 x \cos^4 x dx &= \int \sin^2 x \sin x \cos^4 x dx \qquad \text{Apply Trigonometric Identity } \sin^2 x + \cos^2 x = 1 \text{ for } \sin^2 x\\
\\
\int \sin^3 x \cos^4 x dx &= \int \left( 1 - \cos^2 x \right) \sin x \cos^4 x dx\\
\\
\int \sin^3 x \cos^4 x dx &= \int \left( \cos^4 x - \cos^6 x \right) \sin x dx
\end{aligned}
\end{equation}
$

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

$
\begin{equation}
\begin{aligned}
\int \left( \cos^4 x - \cos^6 x \right) \sin x dx &= \int \left( u^4 - u^6\right) - du\\
\\
\int \left( \cos^4 x - \cos^6 x \right) \sin x dx &= \int - \left( u^4 - u^6\right) du\\
\\
\int \left( \cos^4 x - \cos^6 x \right) \sin x dx &= \int \left( u^6 - u^4 \right) du\\
\\
\int \left( \cos^4 x - \cos^6 x \right) \sin x dx &= \frac{u^{6+1}}{6+1} - \frac{u^{4+1}}{4+1} + c\\
\\
\int \left( \cos^4 x - \cos^6 x \right) \sin x dx &= \frac{u^7}{7} - \frac{u^5}{5} + c\\
\\
\int \left( \cos^4 x - \cos^6 x \right) \sin x dx &= \frac{(\cos x)^7}{7} - \frac{(\cos x)^5}{5} + c\\
\\
\int \left( \cos^4 x - \cos^6 x \right) \sin x dx &= \frac{\cos^7 x }{7} - \frac{\cos^5 x}{5} + c
\end{aligned}
\end{equation}
$