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.