5.5 The Substitution Rule/63

$\int _{0}^{a}x\,\cdot {\sqrt {x^{2}+a^{2}}}dx$

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

${\begin{aligned}{\frac {1}{2}}\int _{a^{2}}^{2a^{2}}{\sqrt {u}}\,\cdot du={\frac {1}{2}}\,\cdot {\frac {a^{{\frac {1}{2}}+1}}{{\frac {1}{2}}+1}}{\bigg |}_{a^{2}}^{2a^{2}}\end{aligned}}$

${\begin{aligned}{\frac {1}{2}}\,\cdot {\frac {2}{3}}\,\cdot \ u^{\frac {3}{2}}{\bigg |}_{a^{2}}^{2a^{2}}={\frac {1}{3}}\left(2a^{2}\right)^{\frac {3}{2}}-{\frac {1}{3}}\left(a^{2}\right)^{\frac {3}{2}}\end{aligned}}$

${\begin{aligned}={\frac {1}{3}}\,\cdot 2^{\frac {3}{2}}\,\cdot {a^{3}}-{\frac {1}{3}}{a^{3}}={\frac {1}{3}}\left(2^{\frac {3}{2}}-1\right){a^{3}}\end{aligned}}$

${\begin{aligned}{\frac {1}{3}}\left(2{\sqrt {2}}-1\right){a^{3}}\end{aligned}}$