5.5 The Substitution Rule/54

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

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

New upper limit: $\pi =({\sqrt {\pi }})^{2}$
New lower limit: $0=(0)^{2}$

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