Answer
$18-9\sqrt 2$
Work Step by Step
$\int_{0}^{3}\frac{x^{3}}{\sqrt (x^{2}+9)}dx$
$let$ $x=3tanθ$
$dx=3sec^{2}θdθ$
$0=3tanθ$
$θ=0$
$3=3tanθ$
$1=tanθ$
$θ=\frac{\pi}{4}$
$\int_{0}^{\frac{\pi}{4}}\frac{27tan^{3}θ}{\sqrt (9tan^{2}θ+9)}(3sec^{2}θ)dθ$
$\int_{0}^{\frac{\pi}{4}}\frac{81tan^{3}θsec^{2}θ}{3\sqrt (tan^{2}θ+1)}dθ$
Through the Pythagorean identity $tan^{2}θ+1=sec^{2}θ$:
$27\int_{0}^{\frac{\pi}{4}}\frac{tan^{3}θsec^{2}θ}{\sqrt sec^{2}θ)}dθ$
$27\int_{0}^{\frac{\pi}{4}}\frac{tan^{3}θsec^{2}θ}{secθ}dθ$
$27\int_{0}^{\frac{\pi}{4}}tan^{3}θsecθdθ$
$27\int_{0}^{\frac{\pi}{4}}tan^{2}θtanθsecθdθ$
Through the Pythagorean identity $tan^{2}θ+1=sec^{2}θ$, $tan^{2}θ=sec^{2}θ-1$:
$27\int_{0}^{\frac{\pi}{4}}(sec^{2}θ-1)tanθsecθdθ$
$let$ $u=secθ$
$du=secθtanθdθ$
$u(0)=1$
$u(\frac{\pi}{4})=\sqrt 2$
$27\int_{1}^{\sqrt 2}(u^{2}-1)du$
$27[\frac{1}{3}u^{3}-u]_{1}^{\sqrt 2}$
$27[\frac{1}{3}(\sqrt 2)^{3}-\sqrt 2-(\frac{1}{3}(1)^{3}-1)]$
$27[\frac{2\sqrt 2}{3}-\sqrt 2-\frac{1}{3}+1]$
$27[\frac{2\sqrt 2}{3}-\sqrt 2+\frac{2}{3}]$
$27[\frac{2\sqrt 2-3\sqrt 2+2}{3}]$
$27[\frac{-\sqrt 2+2}{3}]$
$18-9\sqrt 2$
