Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 56

Determine $\displaystyle \int \sin x \cos x dx$ by four methods.
a.) The substitution $u = \cos x$
If $ u = \cos x$, then $du = -\sin xdx$, so $\sin xdx = - du$. Thus,

$
\begin{equation}
\begin{aligned}
\int \sin x \cos x dx &= \int u \cdot - du\\
\\
\int \sin x \cos x dx &= - \int u du\\
\\
\int \sin x \cos x dx &= -\left( \frac{u^{1+1}}{1+1} \right) + c\\
\\
\int \sin x \cos x dx &= \frac{-u^2}{2} + c \\
\\
\int \sin x \cos x dx &= -\frac{(\cos x)^2}{2} + c\\
\\
\int \sin x \cos x dx &= -\frac{\cos^2 x }{2} + c
\end{aligned}
\end{equation}
$


b.) The substitution $u = \sin x$
If $u = \sin x$, then $du = \cos xdx$. Thus,

$
\begin{equation}
\begin{aligned}
\int \sin x \cos x dx &= \int u du \\
\\
\int \sin x \cos x dx &= \frac{u^{1+1}}{1+1} + c\\
\\
\int \sin x \cos x dx &= \frac{u^2}{2} + c\\
\\
\int \sin x \cos x dx &= \frac{(\sin x)^2}{2} + c\\
\\
\int \sin x \cos x dx &= \frac{\sin^2 x}{2} + c
\end{aligned}
\end{equation}
$


c.) The identity $\sin 2x = 2 \sin x \cos x$
Using the identity $\sin 2x = 2 \sin x \cos x$,

$
\begin{equation}
\begin{aligned}
\int \sin x \cos x dx &= \int \frac{\sin 2x}{2} dx\\
\\
\int \sin x \cos x dx &= \frac{1}{2} \int \sin 2x dx
\end{aligned}
\end{equation}
$

Let $u = 2x$, then $du = 2 dx$, so $\displaystyle dx = \frac{du}{2}$. Thus,

$
\begin{equation}
\begin{aligned}
\int \sin x \cos x dx &= \frac{1}{2} \int \sin u \cdot \frac{du}{2}\\
\\
\int \sin x \cos x dx &= \frac{1}{4} \int \sin u du\\
\\
\int \sin x \cos x dx &= \frac{1}{4} (- \cos u) + c\\
\\
\int \sin x \cos x dx &= \frac{-1}{4} \cos u + c\\
\\
\int \sin x \cos x dx &= \frac{-1}{4} \cos 2 x + c
\end{aligned}
\end{equation}
$


d.) Integration by parts
Let $u = \sin x$, then $du = \cos x dx$ and $v = \sin x$, then $dv = \cos x dx$

$
\begin{equation}
\begin{aligned}
\int \sin x \cos x dx &= uv - \int v du\\
\\
\int \sin x \cos x dx &= \sin x \sin x - \int \sin x \cos x dx\\
\\
\int \sin x \cos x dx &= \sin^2 x - \int \sin x \cos x dx && \text{Combine Like Terms}\\
\\
\int \sin x \cos x dx + \int \sin x \cos x dx &= \sin^2 x\\
\\
2 \int \sin x \cos x dx &= \sin^2 x\\
\\
\int \sin x \cos x dx &= \frac{\sin^2 x}{2} + c
\end{aligned}
\end{equation}
$

Explain the different appearances of the answers.
Although the answers have different appearances, if we simplify each answer by using trigonometric identities, it will have the same result.