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 ${\displaystyle \frac{1}{5}x^{10}+\frac{1}{3}x^6+C}$

 A: You are good except for your last step; $\tan(\theta)$ is not $x^2$, and in particular $\csc$ is not $1/\sec$.
One way to identify what $\tan(\theta)$ is in terms of $x$ is to consider the triangle that justifies the substitution $x=\sec(\theta)$ in the first place. This has a leg of $\sqrt{x^2-1}$, which means the hypotenuse is $x$ and the other leg is $1$. By choosing $u=\sec(\theta)$ you make $\theta$ adjacent to the leg of length $1$ and opposite the leg of length $\sqrt{x^2-1}$. Thus $\tan(\theta)=\sqrt{x^2-1}$. Alternately you could have figured that out by simply using the identity $\sec^2(\theta)=\tan^2(\theta)+1$.
A: How about this:
\begin{align}
u & = \sqrt{x^2-1} \\
u^2 & = x^2 - 1 \\
u^2+1 & = x^2 \\
2u\,du & = 2x\,dx \\
\end{align}
\begin{align}
\int x^3 \sqrt{x^2-1} \, dx & = \int x^2 \sqrt{x^2-1} \Big( x\,dx\Big) \\
& = \int (u^2+1) u \left( \frac 1 2 \, du \right) \\
& = \frac 1 2 \int (u^4 + u^2) \, du \\
& \phantom{{}={}} \text{etc.}
\end{align}
A: 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.
