integral x^3[sqrt(9+x^2)]
i know i am supposed to use trig sub but i am having a very confusing time trying to remember how its done.....
$\displaystyle \int x^3 \sqrt{9+x^2} dx = \int \frac{x^3(9+x^2)^{3/2}}{1}\cdot \frac{1}{9+x^2}dx$
Let,
$\displaystyle t=\tan^{-1} (x/3)$ then $\displaystyle t' = \frac{1}{9+x^2}$
And, $\displaystyle \sec t = \frac{\sqrt{9+x^2}}{3}$ and $\displaystyle \sin t = \frac{x}{3}$
Thus, $\displaystyle 27\sec^3t=(9+x^2)^{3/2}$ and $\displaystyle 27\sin^3t = x^3$
After this substitution we have,
$\displaystyle \int 27^2 \sec^3 t \sin^3 t dt = 729 \int \tan^3 tdt$
Can you do it frum heir?
Let's set $\displaystyle u=9+x^2\implies du=2x\,dx$, this yields
$\displaystyle \begin{aligned}
\int {x^3 \sqrt {9 + x^2 }\,dx} &= \frac{1}
{2}\int {(u - 9)u^{1/2}\,du}\\
&= \frac{1}
{2}\left( {\int {u^{3/2}\,du} - 9\int {u^{1/2}\,du} } \right)\\
&= \frac{1}
{2}\left( {\frac{2}
{5}u^{5/2} - 9 \cdot \frac{2}
{3}u^{3/2} } \right) + c\\
&= \frac{1}
{5}(9 + x^2 )^{5/2} - 3(9 + x^2 )^{3/2} + c
\end{aligned}$
Regards