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