Evaluate the integral: I = integral dx/(sqrt(1+x-x²))
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{1+x-x^2}}. $$
Explanation of Method
Completing the square and applying a sine substitution reduces the integral to $\int d\theta$, giving an arcsine answer that can also be re-expressed as an arctangent using a half-angle-style identity.
- Complete the square for $1+x-x^2$.
- Substitute $x-\tfrac12=\tfrac{\sqrt5}{2}\sin\theta$.
- Simplify to $\int d\theta$.
- Express the result as an arcsine, and note the equivalent arctangent form.
Step 1 · Complete the square
$$ 1+x-x^2 = \frac54-\left(x-\frac12\right)^2. $$
Step 2 · Trigonometric substitution
Let $x-\tfrac12=\dfrac{\sqrt5}{2}\sin\theta$, so $dx=\dfrac{\sqrt5}{2}\cos\theta\,d\theta$ and $\sqrt{1+x-x^2}=\dfrac{\sqrt5}{2}\cos\theta$:
$$ I = \int \frac{\frac{\sqrt5}{2}\cos\theta\,d\theta}{\frac{\sqrt5}{2}\cos\theta} = \int d\theta = \theta + C. $$
Step 3 · Express as arcsine
With $\theta=\arcsin\dfrac{x-1/2}{\sqrt5/2}$:
$$ I = \arcsin\left(\frac{2x-1}{\sqrt5}\right) + C. $$
Step 4 · Equivalent arctangent form
The same antiderivative can be written using the arctangent identity that arises from the alternate ("Euler") substitution $\sqrt{1+x-x^2}=t-x$; this yields the equivalent closed form:
$$ I = -2\arctan\left(\frac{1+\sqrt{1+x-x^2}}{x}\right) + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Completed square | $\tfrac54-(x-\tfrac12)^2$ |
| Arcsine form | $\arcsin\!\left(\tfrac{2x-1}{\sqrt5}\right)$ |
| Arctangent form | $-2\arctan\!\left(\tfrac{1+\sqrt{1+x-x^2}}{x}\right)$ |
Additional Notes
- The domain requires $\tfrac{1-\sqrt5}{2}\leq x\leq\tfrac{1+\sqrt5}{2}$, where $1+x-x^2\geq0$; the arctangent form additionally excludes $x=0$.
- The two forms are both valid antiderivatives (they differ only by an additive constant on their common domain); the arctangent form was verified independently by direct differentiation rather than derived step-by-step from the arcsine form.
- The result was checked by differentiating $-2\arctan\!\left(\tfrac{1+h}{x}\right)$ with $h=\sqrt{1+x-x^2}$: using $h^2+x^2=1+x$, the numerator and denominator of the resulting derivative both reduce to a common factor of $x+2+2h$, leaving exactly $1/h=1/\sqrt{1+x-x^2}$, matching the integrand.
Result
The solution to the integral is: