5.5 The Substitution Rule/54

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

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

${\begin{aligned}\int _{0}^{7}{\sqrt {4+3x}}\,dx=\int _{4}^{25}{\sqrt {u}}\,du\\[2ex]&=\int (du)\sin {(u)}=\int \sin {(u)}du\\[2ex]&=-\cos {(u)}+C\\[2ex]&=-\cos {(\ln {(x)})}+C\end{aligned}}$