Is this an acceptable trig-sub and reversion from trig at the answer? Today I had to take the indefinite integral $\int x^3 \sqrt{x^2-1} \, dx$
My steps: 


*

*$x=\sec\theta$, $dx=\sec\theta\tan\theta\, d\theta$

*$\displaystyle \int \sec^3\theta \sqrt{\sec^2\theta - 1} \, \sec\theta\tan\theta\, d\theta$

*${\displaystyle \int \sec^3\theta \sqrt{\tan^2\theta}\sec\theta\tan\theta d\theta}$

*${\displaystyle \int \sec^3\theta \tan\theta \cdot \sec\theta \tan\theta d\theta}$

*${\displaystyle \int \sec^4\theta \cdot \tan^2\theta d\theta}$

*${\displaystyle \int \sec^2\theta \sec^2\theta\tan^2\theta d\theta}$

*$u=\tan\theta$, $du= \sec^2\theta d\theta$

*${\displaystyle \int u^2(u^2 + 1)}$

*${\displaystyle \int u^4 + u^2}$

*${\displaystyle = \frac{1}{5}u^5 + \frac{1}{3}u^3 + C}$


here is where I'm even more shaky:


*${\displaystyle \frac{1}{5}\tan^5 + \frac{1}{3}\tan^3 + C}$

*since $\tan= \sin/\cos$, $\tan$ also $= \sec/\csc$, and if $\sec\theta=x$, per the substitution, couldn't $\tan$ also be written $x/(1/x)=x^2$, right?

*I replaced all $\tan$s in the final answer with $x^2$, leaving a final answer of ${\displaystyle \frac{1}{5}x^{10}+\frac{1}{3}x^6+C}$


This would have been lots easier with $x=\cosh\theta$:
$$\begin{align}\int x^3\sqrt{x^2-1}dx&=\int\cosh^3\theta\sinh^2\theta\,d\theta\\
&=\int(\sinh^2\theta+1)\sinh^2\theta\cosh\theta\,d\theta\\
&=\frac15\sinh^5\theta+\frac13\sinh^3\theta+C\\
&=\frac15(x^2-1)^{5/2}+\frac13(x^2-1)^{3/2}+C\\
&=\left(\frac15x^2-\frac15+\frac13\right)(x^2-1)^{3/2}+C\\
&=\frac1{15}(3x^2+2)(x^2-1)^{3/2}+C\end{align}$$
But I don't know if hyperbolic substitution counts as trigonometric substitution in this context.
