Evaluate the integral: I = integral sqrt(a²-x²) dx
Evaluate the integral: $$ I = \int \sqrt{a^2-x^2}\,dx. $$
Explanation of Method
The trigonometric substitution $x=a\sin\theta$ removes the square root entirely, turning the integral into a power-reduction problem in $\cos^2\theta$.
- Substitute $x=a\sin\theta$.
- Simplify the square root using $1-\sin^2\theta=\cos^2\theta$.
- Apply the power-reduction identity to $\cos^2\theta$.
- Substitute back using $\theta=\arcsin(x/a)$ and $\sin(2\arcsin u)=2u\sqrt{1-u^2}$.
Step 1 · Trigonometric substitution
Let $x=a\sin\theta$, so $dx=a\cos\theta\,d\theta$, and:
$$ \sqrt{a^2-x^2} = \sqrt{a^2-a^2\sin^2\theta} = a\cos\theta. $$
Substituting into the integral:
$$ I = \int a\cos\theta\cdot a\cos\theta\,d\theta = a^2\int \cos^2\theta\,d\theta. $$
Step 2 · Power-reduction identity
Using $\cos^2\theta=\dfrac{1+\cos2\theta}{2}$:
$$ I = \frac{a^2}{2}\int d\theta + \frac{a^2}{2}\int \cos2\theta\,d\theta = \frac{a^2}{2}\theta + \frac{a^2}{4}\sin2\theta + C. $$
Step 3 · Substitute back
With $\theta=\arcsin(x/a)$ and $\sin(2\arcsin u)=2u\sqrt{1-u^2}$, $u=x/a$:
$$ \sin2\theta = 2\cdot\frac{x}{a}\sqrt{1-\frac{x^2}{a^2}} = \frac{2x}{a^2}\sqrt{a^2-x^2}. $$
So the second term becomes $\dfrac{a^2}{4}\cdot\dfrac{2x}{a^2}\sqrt{a^2-x^2}=\dfrac{x}{2}\sqrt{a^2-x^2}$.
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $x=a\sin\theta$ |
| Reduced integral | $a^2\int\cos^2\theta\,d\theta$ |
| $\theta$ back-substitution | $\arcsin(x/a)$ |
Additional Notes
- The domain requires $-a\leq x\leq a$, where $\sqrt{a^2-x^2}$ is real and $\arcsin(x/a)$ is defined.
- This is one of the three classic Euler/trigonometric-substitution templates for radicals: $a^2-x^2$ calls for $x=a\sin\theta$, while $x^2\pm a^2$ (Problem 182) instead call for secant/tangent substitutions or integration by parts.
- The result can be checked by differentiating $\tfrac{a^2}{2}\arcsin(x/a)+\tfrac{x}{2}\sqrt{a^2-x^2}$ and confirming it returns $\sqrt{a^2-x^2}$.
Result
The solution to the integral is: