Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 38

Find the definite integral $\displaystyle \int^{\sqrt{\pi}}_0 x \cos (x^2) dx$

Let $u = x^2$, then $du = 2x dx$, so $\displaystyle xdx = \frac{du}{2}$. When $x = 0, u = 0$ and when $x = \sqrt{\pi}, u = \pi$. Thus,


$
\begin{equation}
\begin{aligned}

\int^{\sqrt{\pi}}_0 x \cos (x^2) dx =& \int^{\sqrt{\pi}}_0 \cos (x^2) x dx
\\
\\
\int^{\sqrt{\pi}}_0 x \cos (x^2) dx =& \int^{\sqrt{\pi}}_0 \cos u \frac{du}{2}
\\
\\
\int^{\sqrt{\pi}}_0 x \cos (x^2) dx =& \frac{1}{2} \int^{\sqrt{\pi}}_0 \cos u du
\\
\\
\int^{\sqrt{\pi}}_0 x \cos (x^2) dx =& \left. \frac{1}{2} \sin u \right|^{\sqrt{\pi}}_0
\\
\\
\int^{\sqrt{\pi}}_0 x \cos (x^2) dx =& \frac{\sin (\pi)}{2} - \frac{\sin (0)}{2}
\\
\\
\int^{\sqrt{\pi}}_0 x \cos (x^2) dx =& 0 - 0
\\
\\
\int^{\sqrt{\pi}}_0 x \cos (x^2) dx =& 0


\end{aligned}
\end{equation}
$