Evaluate the integral: I = integral (x dx)/(sqrt(1 - x²))
Evaluate the integral: $$ I = \int \frac{x\,dx}{\sqrt{1 - x^2}}. $$
Explanation of Method
With $u=1-x^2$, the numerator $x\,dx$ matches (up to a constant) the derivative $du=-2x\,dx$, converting the square root into $\sqrt{u}$ and reducing the integral to a routine negative-fractional-power antiderivative.
- Substitute $u$ for the expression under the root.
- Match $x\,dx$ to a constant multiple of $du$.
- Integrate the resulting fractional power of $u$.
- Substitute back and add the constant of integration.
Step 1 · Set up the substitution
To solve this integral, we use the substitution method. Observing the denominator, we choose:
$$ u = 1 - x^2. $$
Now, differentiate both sides with respect to $x$:
$$ \frac{du}{dx} = -2x. $$
Rearranging, we get:
$$ du = -2x\,dx. $$
This allows us to express $x\,dx$ in terms of $du$:
$$ x\,dx = -\frac{du}{2}. $$
Step 2 · Rewrite and integrate
Substituting into the integral:
$$ I = \int \frac{x\,dx}{\sqrt{1-x^2}} = \int \frac{-\frac{du}{2}}{\sqrt{u}}. $$
Factor out the constant:
$$ I = -\frac{1}{2}\int u^{-1/2}\,du. $$
The integral of $u^{-1/2}$ is:
$$ \int u^{-1/2}\,du = 2u^{1/2}. $$
Thus, we obtain:
$$ I = -\frac{1}{2}\times 2u^{1/2} + C = -\sqrt{u} + C. $$
Step 3 · Substitute back
Finally, substituting back $u = 1-x^2$, we get:
$$ I = -\sqrt{1-x^2} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = 1-x^2$ |
| $x\,dx$ in terms of $du$ | $-\tfrac12 du$ |
| Antiderivative | $-\sqrt{u}$ |
Additional Notes
- The domain of $I$ requires $-1 < x < 1$, so that $1-x^2>0$ under the square root in the denominator.
- This is one of the simplest possible substitutions: the numerator is exactly proportional to the derivative of the radicand, so no extra rewriting is needed beyond the constant factor.
- The result can be checked by differentiating $-\sqrt{1-x^2}$ with the chain rule and confirming it returns $\dfrac{x}{\sqrt{1-x^2}}$.
Result
The solution to the integral is: