How do I integrate x^3(x^2-1)^3/2

Question

fibonaccichick666 · Answer

let me just double check, this is your question yes? $$\int x^3(x^2-1)^{3/2}dx$$

anonymous · Answer

Yes

anonymous · Answer

Trig sub is my first instinct. Set $\sec^{-1}x=u$ so that $dx=\sec u\tan u\,du$, then
$$\begin{align*}\int x^3(x^2-1)^{3/2}\,dx&=\int\sec^3u(\sec^2u-1)^{3/2}\sec u\tan u\,du\\
&=\int\sec^3u(\tan^2u)^{3/2}\sec u\tan u\,du\\
&=\int\sec^3u\tan^3u\sec u\tan u\,du\\
&=\int\sec^2u\tan^4u\sec^2u\,du\\
&=\int(1+\tan^2u)\tan^4u\sec^2u\,du\\
&\text{setting }t=\tan u\text{ and }dt=\sec^2u\,du\\
&=\int(1+t^2)t^4\,dt
\end{align*}$$

anonymous · Answer

put $$x^2-1=t,x^2=1+t,2 x~dx=dt$$
$$I=\int\limits x^2\left( x^2-1 \right)^{\frac{ 3 }{ 2 }}~x~dx=\frac{ 1 }{ 2 }\int\limits \left( 1+t \right)t ^{\frac{ 3 }{ 2 }} dt$$
$$=\frac{ 1 }{ 2 }\int\limits \left[ t ^{\frac{ 3 }{ 2 }}+t ^{1+\frac{ 3 }{ 2 }} \right]dt=\frac{ 1 }{ 2 }\int\limits \left[ t ^{\frac{ 3 }{ 2 }}+t ^{\frac{ 5 }{ 2 }} \right]dt+c$$
?