Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 32

Find the indefinite integral $\displaystyle \int \frac{\sin \sqrt{x}}{\sqrt{x}} dx$. Illustrate by graphing both function and its anti-derivative (take $C = 0$).

If we let $u = \sqrt{x}$, then $\displaystyle du = \frac{1}{2 \sqrt{x}} dx$, so $\displaystyle \frac{1}{\sqrt{x}} dx = 2 du$. And


$
\begin{equation}
\begin{aligned}

\int \frac{\sin \sqrt{x}}{\sqrt{x}} dx =& \int \sin \sqrt{x} \frac{1}{\sqrt{x}} dx
\\
\\
\int \frac{\sin \sqrt{x}}{\sqrt{x}} dx =& \int \sin u 2 du
\\
\\
\int \frac{\sin \sqrt{x}}{\sqrt{x}} dx =& 2 \int \sin u du
\\
\\
\int \frac{\sin \sqrt{x}}{\sqrt{x}} dx =& 2 (- \cos u) + C
\\
\\
\int \frac{\sin \sqrt{x}}{\sqrt{x}} dx =& -2 \cos u + C
\\
\\
\int \frac{\sin \sqrt{x}}{\sqrt{x}} dx =& -2 \cos \sqrt{x} + C

\end{aligned}
\end{equation}
$


The graph of $\displaystyle y = \frac{\sin \sqrt{x}}{\sqrt{x}}$








The graph of anti-derivative $y = - 2 \cos \sqrt{x}$