Evaluate the integral: I = integral dx/((x²-1) sqrt(1-x²))
Evaluate the integral: $$ I = \int \frac{dx}{\left(x^2-1\right)\sqrt{1-x^2}}. $$
Explanation of Method
On the domain $-1<x<1$, $x^2-1=-(1-x^2)$, so the integrand simplifies to $-1/(1-x^2)^{3/2}$. The trigonometric substitution $x=\sin\theta$ then reduces the integral to $-\int\sec^2\theta\,d\theta$.
- Simplify using $x^2-1=-(1-x^2)$ on the domain.
- Substitute $x=\sin\theta$.
- Integrate $-\sec^2\theta$.
- Convert back to $x$.
Step 1 · Simplify
The integral requires $-1<x<1$ (so $1-x^2>0$) and $x\neq\pm1$; on this domain $x^2-1<0$, so:
$$ \frac1{(x^2-1)\sqrt{1-x^2}} = \frac{-1}{(1-x^2)^{3/2}} \quad\Rightarrow\quad I = -\int\frac{dx}{(1-x^2)^{3/2}}. $$
Step 2 · Trigonometric substitution
Let $x=\sin\theta$, $-\tfrac\pi2<\theta<\tfrac\pi2$, so $dx=\cos\theta\,d\theta$ and $\sqrt{1-x^2}=\cos\theta$:
$$ -\int\frac{\cos\theta\,d\theta}{\cos^3\theta} = -\int\sec^2\theta\,d\theta = -\tan\theta+C. $$
Step 3 · Convert back to $x$
Since $\sin\theta=x$ and $\cos\theta=\sqrt{1-x^2}$, $\tan\theta=\dfrac x{\sqrt{1-x^2}}$:
$$ I = -\frac x{\sqrt{1-x^2}} + C. $$
Summary Table
| Substitution | Relation |
|---|---|
| $x=\sin\theta$ | $dx=\cos\theta\,d\theta$ |
| $\sqrt{1-x^2}$ | $\cos\theta$ |
Additional Notes
- The domain is $-1<x<1$.
- The result was verified by differentiation: $\dfrac{d}{dx}\left[-\dfrac x{\sqrt{1-x^2}}\right] = -\dfrac1{(1-x^2)^{3/2}} = \dfrac1{(x^2-1)\sqrt{1-x^2}}$, matching the integrand exactly.
Result
The solution to the integral is: