5.5 The Substitution Rule/54

$\int _{0}^{\sqrt {\pi }}x\cos {(x^{2})}\,dx$

${\begin{aligned}u&=x^{2}\\[2ex]du&=2xdx\\[2ex]{\frac {1}{2}}du&=xdx\\[2ex]\end{aligned}}$

${\begin{aligned}\int _{0}^{\sqrt {\pi }}x\cos {(x^{2})}\,dx&=\int _{0}^{\sqrt {\pi }}(xdx)\cos {(x^{2})}=&=\int ({\frac {1}{2}}du)\cos {(u)}={\frac {1}{2}}\int \cos {(u)}du\\[2ex]&={\frac {1}{2}}\sin {(u)}+C\\[2ex]&={\frac {1}{2}}\sin {(x^{2})}+C\\[2ex]\end{aligned}}$