Evaluate the integral: I = integral dx/((1+x²) sqrt(1-x²))
Evaluate the integral: $$ I = \int \frac{dx}{(1+x^2)\sqrt{1-x^2}}. $$
Explanation of Method
The substitution $x=\sin\theta$ collapses the radical, reducing the integral to $\int d\theta/(1+\sin^2\theta)$, which converts to a rational function of $\tan\theta$ via division by $\cos^2\theta$.
- Substitute $x=\sin\theta$.
- Convert to tangent form.
- Integrate.
- Back-substitute.
Step 1 · Substitute
Let $x=\sin\theta$, so $dx=\cos\theta\,d\theta$ and $\sqrt{1-x^2}=\cos\theta$:
$$ I = \int\frac{d\theta}{1+\sin^2\theta}. $$
Step 2 · Convert to tangent form
Dividing numerator and denominator by $\cos^2\theta$ and using $\sec^2\theta=1+\tan^2\theta$:
$$ I = \int\frac{\sec^2\theta\,d\theta}{1+2\tan^2\theta}. $$
Step 3 · Integrate
Let $t=\tan\theta$, $dt=\sec^2\theta\,d\theta$:
$$ I = \int\frac{dt}{1+2t^2} = \frac1{\sqrt2}\arctan\left(\sqrt2t\right) + C. $$
Step 4 · Back-substitute
With $t=\tan\theta=x/\sqrt{1-x^2}$:
$$ I = \frac1{\sqrt2}\arctan\left(\frac{\sqrt2x}{\sqrt{1-x^2}}\right) + C. $$
Summary Table
| Substitution | Effect |
|---|---|
| $x=\sin\theta$ | reduces to $\int d\theta/(1+\sin^2\theta)$ |
| $t=\tan\theta$ | reduces to $\int dt/(1+2t^2)$ |
Additional Notes
- The domain is $-1<x<1$.
- The result was verified by differentiation at $x=1/2$ and $x=-1/3$: the derivative matches the integrand $1/[(1+x^2)\sqrt{1-x^2}]$ exactly at both points.
Result
The solution to the integral is: