Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 58

Find the area of the region bounded by the curves $y = \sin ^3 x$ and $y = \cos^3 x$ from $\displaystyle \frac{\pi}{4} \leq x \leq \frac{5\pi}{4}$



By using vertical strips,

$
\begin{equation}
\begin{aligned}
A &= \int^b_a \left( y_{\text{upper}} - y_{\text{lower}} \right) dx\\
\\
A &= \int^{5\pi/4}_{\pi/4} \left( \sin^3 x - \cos^3 x \right) dx\\
\\
A &= \int^{5\pi/4}_{\pi/4} \sin^3 x dx - \int^{5\pi/4}_{\pi/4} \cos^3 x dx
\end{aligned}
\end{equation}
$


For the first term,

$
\begin{equation}
\begin{aligned}
\int \sin^3 x dx &= \int \sin x \cdot \sin ^2 x dx && \text{, recall that } \sin ^2 x = 1 - \cos^2 x\\
\\
&= \int \sin x \left( 1 - \cos^2 x \right) dx = \int \left( \sin x - \sin x \cos^2 x \right) dx\\
\\
&= \int \sin x dx - \int \sin x \cos^2 x dx\\
\\
&= -\cos x - \int \cos^2 x dx
\end{aligned}
\end{equation}
$


To evaluate $\displaystyle \int \sin x \cos^2 x dx$, we let $u = \cos x$, then so
$du = -\sin x dx$

Thus,

$
\begin{equation}
\begin{aligned}
\int \sin x \cos^2 x dx &= \int u^2 (-du) = - \int u^2 du\\
\\
&= -\frac{u^3}{3} = -\frac{\cos^3x}{3}
\end{aligned}
\end{equation}
$


Hence,
$\displaystyle \int \sin^3 x dx = - \cos x + \frac{\cos^3 x}{3}$

For the term $\displaystyle \int \cos^3 x dx$

$
\begin{equation}
\begin{aligned}
\int \cos^3 x &= \int \cos x \cdot \cos^3 x dx \qquad \text{, recall that } \cos^2x = 1 - \sin^2 x\\
\\
&= \int \cos x (1-\sin^2 x) dx = \int \cos x dx - \int \cos x \sin^2 x dx\\
\\
&= \sin x - \int \cos x \sin^2 x dx
\end{aligned}
\end{equation}
$


To evaluate the right term, we let $u = \sin x$, so $du = \cos x dx$
$\displaystyle \int \cos x \sin^2 x dx = \int u^2 du = \frac{u^3}{3} = \frac{\sin^3 x}{3}$
Hence,
$\displaystyle \int \cos^3 x dx = \sin x - \frac{\sin^3 x}{3}$

Therefore, from the previous equation

$
\begin{equation}
\begin{aligned}
A &= \int \sin^3 x dx - \int \cos^3 x dx\\
\\
A &= - \cos x + \frac{\cos^3 x}{3} - \left[ \sin x - \frac{\sin^3 x}{3} \right]\\
\\
A &= \frac{1}{3} \left( \sin^3 x + \cos^3 x \right) - \sin x - \cos x
\end{aligned}
\end{equation}
$



Evaluating $\displaystyle x = \frac{\pi}{4}$ to $\displaystyle x = \frac{5\pi}{4}$


$
\begin{equation}
\begin{aligned}
A &= \frac{1}{3} \left[ \left( -\frac{\sqrt{2}}{2} \right)^3 + \left( -\frac{\sqrt{2}}{2} \right)^3 \right] - \left( -\frac{\sqrt{2}}{2} \right)^3 - \left( -\frac{\sqrt{2}}{2} \right)^3 -
\left[ \frac{1}{3} \left[ \left( \frac{\sqrt{2}}{2} \right)^3 + \left( \frac{\sqrt{2}}{2} \right)^3 \right] - \left( \frac{\sqrt{2}}{2} \right)^3 - \left( \frac{\sqrt{2}}{2} \right)^3 \right]\\
\\

A &= \frac{5 \sqrt{2}}{3} \text{ square units}
\end{aligned}
\end{equation}
$