Evaluate the integral: I = integral dx/(x^6sqrt(x²-1))
Evaluate the integral: $$ I = \int \frac{dx}{x^6\sqrt{x^2-1}}. $$
Explanation of Method
The trig substitution $x=\sec\theta$ collapses $\sqrt{x^2-1}=\tan\theta$ against the $\sec\theta\tan\theta\,d\theta$ from $dx$, reducing the integral to $\int\cos^5\theta\,d\theta$, a standard odd-power cosine integral.
- Substitute $x=\sec\theta$.
- Integrate $\cos^5\theta$ via $u=\sin\theta$.
- Convert back to $x$.
Step 1 · Substitute
Let $x=\sec\theta$, $0\leq\theta<\tfrac\pi2$, so $dx=\sec\theta\tan\theta\,d\theta$ and $\sqrt{x^2-1}=\tan\theta$:
$$ I = \int\frac{\sec\theta\tan\theta\,d\theta}{\sec^6\theta\cdot\tan\theta} = \int\sec^{-5}\theta\,d\theta = \int\cos^5\theta\,d\theta. $$
Step 2 · Integrate $\cos^5\theta$
Write $\cos^5\theta=(1-\sin^2\theta)^2\cos\theta$ and let $u=\sin\theta$, $du=\cos\theta\,d\theta$:
$$ \int(1-u^2)^2\,du = \int\left(1-2u^2+u^4\right)du = u-\frac{2u^3}3+\frac{u^5}5+C. $$
Step 3 · Convert back to $x$
With $\cos\theta=1/x$, $\sin\theta=\sqrt{x^2-1}/x$:
$$ I = \frac{\sqrt{x^2-1}}x - \frac23\cdot\frac{(x^2-1)^{3/2}}{x^3} + \frac15\cdot\frac{(x^2-1)^{5/2}}{x^5} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| $\sin\theta$ | $\sqrt{x^2-1}/x$ |
| $\cos\theta$ | $1/x$ |
Additional Notes
- The domain is $x>1$ (or $x<-1$ by symmetry).
- The result was verified symbolically: differentiating the boxed expression and simplifying gives exactly $1/[x^6\sqrt{x^2-1}]$, confirmed with a numeric check at $x=2$.
Result
The solution to the integral is: