Integral of √(1−x²): Trig Substitution and the Semicircle

The indefinite integral of $$\sqrt{1-x^2}$$ is $$\frac{1}{2}\left(x\sqrt{1-x^2} + \arcsin x\right) + C$$. However, you can solve it using the trigonometric substitution $$x = \sin\theta$$. Over the definite interval $$[-1, 1]$$, the integral equals $$\frac{\pi}{2}$$. This value represents the exact area. It is the area of the upper half of a unit circle with radius 1.

The integral of sqrt(1-x^2) is a standard first example of trigonometric substitution. It appears in calculus. This integral bridges algebraic integration and basic circle geometry. Standard algebraic techniques, like basic substitution, usually fail here. This is because there is no outer $$x$$ factor to match $$du$$. However, trigonometric identities let you rewrite the square root. You can rewrite it as a single squared term. Below, we provide the full step-by-step derivation. We use sine substitution. Also, we include a geometric proof. This proof connects the antiderivative to sectors and triangles. Furthermore, we show an alternative integration by parts method. Finally, we discuss practical applications across shifted circles and ellipses.

How do you integrate √(1-x²) using trig substitution?

Integrating $$\sqrt{1-x^2}$$ requires a conversion. First, convert the radical expression into a trigonometric function. Use the Pythagorean identity $$1 - \sin^2\theta = \cos^2\theta$$. This entirely eliminates the square root.

Step 1: Set up the trigonometric substitution

Let $$x = \sin\theta$$. Taking the differential of both sides gives:

Therefore, $$x = \sin\theta$$. So, $$dx = \cos\theta\,d\theta$$. Also, $$-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$$.

Restricting $$\theta$$ to the interval $$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$ is a crucial requirement for two main reasons:

• First, it ensures the sine function is one-to-one. Therefore, you can uniquely write $$\theta = \arcsin x$$.
• This guarantees that $$\cos\theta \ge 0$$ on this interval. Therefore, you can simplify $$\sqrt{\cos^2\theta}$$ directly as $$\cos\theta$$. You do not need absolute value bars.

Substituting $$x = \sin\theta$$ and $$dx = \cos\theta\,d\theta$$ into the integral gives:

$$\int \sqrt{1 - x^2}\,dx = \int \sqrt{1 - \sin^2\theta}\cdot \cos\theta\,d\theta$$. You can simplify this to $$\int \sqrt{\cos^2\theta}\cdot \cos\theta\,d\theta$$. Therefore, this equals $$\int \cos^2\theta\,d\theta$$.

A common mistake at this stage is forgetting to substitute $$dx$$. Replacing $$dx$$ with $$\cos\theta\,d\theta$$ is what produces the integrand $$\cos^2\theta$$ rather than just $$\cos\theta$$.

Step 2: Integrate cos²θ using the power-reduction identity

To integrate $$\cos^2\theta$$, apply a power-reduction identity. Use $$\cos^2\theta = \frac{1 + \cos 2\theta}{2}$$. Also, for more information, see the guide on integral of sin²x.

$$\begin{aligned}\int \cos^2\theta\,d\theta &= \int \frac{1 + \cos 2\theta}{2}\,d\theta &= \frac{1}{2} \int (1 + \cos 2\theta)\,d\theta &= \frac{1}{2}\left(\theta + \frac{\sin 2\theta}{2}\right) + C &= \frac{\theta}{2} + \frac{\sin 2\theta}{4} + C\end{aligned}$$

To express this result back in terms of $$x$$, apply the double-angle identity $$\sin 2\theta = 2\sin\theta\cos\theta$$. This expands the second term.

$$\frac{\theta}{2} + \frac{2\sin\theta\cos\theta}{4} + C = \frac{\theta}{2} + \frac{\sin\theta\cos\theta}{2} + C$$

Step 3: Convert back to the original variable x

Now substitute back to the variable $$x$$ using the relations established in Step 1:

• Since $$x = \sin\theta$$, taking the inverse sine gives $$\theta = \arcsin x$$.
• Since $$x = \sin\theta$$, the expression $$\cos\theta$$ equals $$\sqrt{1 - x^2}$$.

Substituting these into the simplified trigonometric result gives the final antiderivative:

$$\int \sqrt{1-x^2}\,dx = \frac{\arcsin x}{2} + \frac{x\sqrt{1-x^2}}{2} + C = \frac{1}{2}\left(x\sqrt{1-x^2} + \arcsin x\right) + C$$

If you find it difficult to remember how to evaluate $$\cos\theta$$, sketch a right triangle with angle $$\theta$$. Set the opposite side to $$x$$ and the hypotenuse to $$1$$. By the Pythagorean theorem, the adjacent side length is $$\sqrt{1^2 - x^2} = \sqrt{1 - x^2}$$. From this triangle, $$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \sqrt{1 - x^2}$$.

How can you check the answer by differentiating?

You can verify the antiderivative. Just take its derivative with respect to $$x$$. You will find that $$\frac{1}{2}\left(x\sqrt{1-x^2} + \arcsin x\right)$$ is correct.

Differentiating the term $$x\sqrt{1-x^2}$$ using the product rule yields:

$$\frac{d}{dx}\left(x\sqrt{1-x^2}\right)$$ is $$(1)\sqrt{1-x^2} + x \cdot \frac{-2x}{2\sqrt{1-x^2}}$$. You can simplify this expression. As a result, it becomes $$\sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}$$.

The derivative of arcsin x is $$\frac{1}{\sqrt{1-x^2}}$$. Therefore, combining the derivatives of both terms inside the brackets gives:

$$\frac{d}{dx}\left(x\sqrt{1-x^2} + \arcsin x\right) = \sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}} + \frac{1}{\sqrt{1-x^2}}$$

Combine the two fractional terms over a common denominator:

$$-\frac{x^2}{\sqrt{1-x^2}} + \frac{1}{\sqrt{1-x^2}} = \frac{1 - x^2}{\sqrt{1-x^2}} = \sqrt{1-x^2}$$

Substituting this back into the total expression gives:

$$\frac{d}{dx}\left(x\sqrt{1-x^2} + \arcsin x\right) = \sqrt{1-x^2} + \sqrt{1-x^2} = 2\sqrt{1-x^2}$$

We multiply by the factor of $$\frac{1}{2}$$ in front. This yields exactly $$\sqrt{1-x^2}$$. Thus, the antiderivative formula is correct.

Why does the formula split into a triangle and a circular sector?

The equation $$y = \sqrt{1 - x^2}$$ describes the top half of the unit circle centered at the origin, $$x^2 + y^2 = 1$$. For any value $$x$$ between $$0$$ and $$1$$, the definite integral $$\int_0^x \sqrt{1 - t^2}\,dt$$ calculates the total area under the upper semicircle from $$t = 0$$ to $$t = x$$.

Geometrically, this total area breaks into two distinct shapes:

• A right triangle: This triangle has vertices at the origin $$(0,0)$$ and the point on the x-axis $$(x, 0)$$. Also, it has a vertex at the point on the circle $$(x, \sqrt{1-x^2})$$. Its base is $$x$$ and its height is $$\sqrt{1-x^2}$$. Therefore, its area is $$\frac{1}{2}x\sqrt{1-x^2}$$.
• A circular sector: This is a slice of the circle. The vertical y-axis and the radial line from $$(0,0)$$ to $$(x, \sqrt{1-x^2})$$ bound it. The central angle with the y-axis satisfies $$\sin\theta = x$$. So, $$\theta = \arcsin x$$. In a unit circle, the radius is 1. Therefore, the area of a sector with central angle $$\theta$$ radians is $$\frac{1}{2}r^2\theta = \frac{1}{2}\arcsin x$$.

Add the area of the right triangle and the circular sector. This gives you $$\frac{1}{2}x\sqrt{1-x^2} + \frac{1}{2}\arcsin x$$. The algebraic terms in the antiderivative are not arbitrary calculus artifacts. Instead, they directly map to the circle’s geometry.

How do you integrate √(1-x²) using integration by parts?

Do you prefer to avoid trigonometric substitution? Integration by parts offers an alternative. This method provides an algebraic derivation. First, set the integral as $$I = \int \sqrt{1-x^2}\,dx$$.

Choose the parts as follows:

• $$u = \sqrt{1-x^2} \implies du = \frac{-x}{\sqrt{1-x^2}}\,dx$$
• $$dv = dx \implies v = x$$

Applying the integration by parts formula $$\int u\,dv = uv - \int v\,du$$ gives:

$$I = x\sqrt{1-x^2} - \int x \cdot \left(\frac{-x}{\sqrt{1-x^2}}\right)dx$$. So, this simplifies to $$x\sqrt{1-x^2} + \int \frac{x^2}{\sqrt{1-x^2}}\,dx$$.

Solving the remaining integral

Rewrite the numerator $$x^2$$ as $$1 - (1 - x^2)$$. Then split the fraction into two separate terms.

$$\begin{aligned}\int \frac{x^2}{\sqrt{1-x^2}}\,dx &= \int \frac{1 - (1-x^2)}{\sqrt{1-x^2}}\,dx &= \int \frac{1}{\sqrt{1-x^2}}\,dx - \int \frac{1-x^2}{\sqrt{1-x^2}}\,dx &= \arcsin x - \int \sqrt{1-x^2}\,dx\end{aligned}$$

Notice that the original integral $$I = \int \sqrt{1-x^2}\,dx$$ has reappeared on the right side of the equation:

$$I = x\sqrt{1-x^2} + \arcsin x - I$$

Isolating the integral I

Add $$I$$ to both sides of the equation:

$$2I = x\sqrt{1-x^2} + \arcsin x$$

Divide by 2 and add the constant of integration $$C$$:

$$I = \frac{1}{2}\left(x\sqrt{1-x^2} + \arcsin x\right) + C$$

Both trigonometric substitution and integration by parts reach the same antiderivative. However, trigonometric substitution is usually better in introductory courses. This is because it avoids an algebraic trick. The trick involves adding and subtracting 1 in the numerator.

How do geometric shortcuts solve definite integrals of semicircles?

The equation $$y = \sqrt{1-x^2}$$ represents a semicircle of radius 1. Because of this, you can evaluate definite integrals instantly. Just use standard geometry formulas. You do not need to compute antiderivatives. This applies over specific symmetry intervals.

• Full semicircle ($[-1, 1]$): $$\int_{-1}^{1} \sqrt{1-x^2}\,dx = \frac{\pi (1)^2}{2} = \frac{\pi}{2}$$
• Quarter circle ($[0, 1]$ or $[-1, 0]$): $$\int_0^1 \sqrt{1-x^2}\,dx = \frac{\pi (1)^2}{4} = \frac{\pi}{4}$$

You can save time on exam questions by recognizing circular geometry. Before you set up a substitution for a definite integral containing $$\sqrt{a^2 - x^2}$$, sketch the region. Check if the bounds align with a quarter-disk or half-disk.

How do you evaluate integrals with general radius a or shifted centers?

When the term under the square root is $$\sqrt{a^2 - x^2}$$ for a general constant $$a > 0$$, use the trigonometric substitution $$x = a\sin\theta$$ with $$dx = a\cos\theta\,d\theta$$.

The antiderivative formula for a general radius $$a$$ is:

$$\int \sqrt{a^2 - x^2}\,dx = \frac{x\sqrt{a^2 - x^2}}{2} + \frac{a^2}{2}\arcsin\left(\frac{x}{a}\right) + C$$

For the definite integral across the full diameter $$[-a, a]$$, the geometric area formula yields:

$$\int_{-a}^{a} \sqrt{a^2 - x^2}\,dx = \frac{\pi a^2}{2}$$

Completing the square for shifted circles

If the integrand contains a quadratic expression, you must complete the square. For example, use this method for $$\sqrt{bx - x^2}$$. This brings it into the standard $$\sqrt{a^2 - u^2}$$ form.

For example, to evaluate $$\int \sqrt{2x - x^2}\,dx$$:

1. Rewrite $$2x - x^2$$ as $$-(x^2 - 2x) = -((x-1)^2 - 1) = 1 - (x-1)^2$$.
2. Substitute $$u = x - 1$$, which gives $$du = dx$$.
3. The integral becomes $$\int \sqrt{1 - u^2}\,du$$, which matches the unit circle integral in terms of $$u$$.

Worked examples step-by-step

Example 1: Definite integral over a quarter disk

Compute $$\int_0^1 \sqrt{1-x^2}\,dx$$ by changing substitution limits.

Using $$x = \sin\theta$$ and $$dx = \cos\theta\,d\theta$$:

• When $$x = 0$$, $$\sin\theta = 0 \implies \theta = 0$$.
• When $$x = 1$$, $$\sin\theta = 1 \implies \theta = \frac{\pi}{2}$$.

Transforming the limits into $$\theta$$ allows direct evaluation without converting back to $$x$$:

$$\int_0^{\pi/2} \cos^2\theta\,d\theta = \left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_0^{\pi/2} = \left(\frac{\pi/2}{2} + \frac{\sin\pi}{4}\right) - (0 + 0) = \frac{\pi}{4}$$

This result matches the geometric area of a quarter-disk of radius 1.

Example 2: Integration over an arbitrary interval [0, 1/2]

Evaluate $$\int_0^{1/2} \sqrt{1-x^2}\,dx$$ using the antiderivative formula.

Substitute the upper limit $$x = \frac{1}{2}$$ and lower limit $$x = 0$$ into $$\frac{1}{2}\left(x\sqrt{1-x^2} + \arcsin x\right)$$:

$$\text{At } x = \frac{1}{2}: \quad \frac{1}{2}\left(\frac{1}{2}\sqrt{1 - \frac{1}{4}} + \arcsin\frac{1}{2}\right) = \frac{1}{2}\left(\frac{\sqrt{3}}{4} + \frac{\pi}{6}\right) = \frac{\sqrt{3}}{8} + \frac{\pi}{12}$$

$$\text{At } x = 0: \quad \frac{1}{2}\left(0 + \arcsin 0\right) = 0$$

Subtracting lower from upper gives $$\frac{\sqrt{3}}{8} + \frac{\pi}{12}$$. In terms of geometry, $$\frac{\sqrt{3}}{8}$$ represents the triangle portion and $$\frac{\pi}{12}$$ represents the sector portion.

Example 3: Radius 3 quarter-disk

Evaluate $$\int_0^3 \sqrt{9 - x^2}\,dx$$.

Since this represents a quarter circle of radius $$a = 3$$, applying the geometric shortcut yields:

$$\int_0^3 \sqrt{9 - x^2}\,dx = \frac{\pi (3)^2}{4} = \frac{9\pi}{4}$$

Example 4: Shifted circle over full semicircle

Evaluate $$\int_0^2 \sqrt{2x - x^2}\,dx$$.

Complete the square: $$2x - x^2 = 1 - (x-1)^2$$. Let $$u = x - 1$$, so $$du = dx$$. When $$x = 0$$, $$u = -1$$; when $$x = 2$$, $$u = 1$$.

$$\int_0^2 \sqrt{2x - x^2}\,dx = \int_{-1}^{1} \sqrt{1 - u^2}\,du = \frac{\pi}{2}$$

Example 5: Proving the area formula of an ellipse

Find the total area enclosed by the ellipse $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$.

Solving for $$y$$ in the upper half gives $$y = \frac{2}{3}\sqrt{9 - x^2}$$. By four-fold symmetry, the total area is four times the first-quadrant area from $$x = 0$$ to $$x = 3$$:

$$\text{Area} = 4 \int_0^3 \frac{2}{3}\sqrt{9 - x^2}\,dx = \frac{8}{3} \int_0^3 \sqrt{9 - x^2}\,dx$$

Using the result from Example 3 ($$\frac{9\pi}{4}$$):

$$\text{Area} = \frac{8}{3} \cdot \frac{9\pi}{4} = 6\pi$$

This matches the general ellipse area formula $$\text{Area} = \pi a b = \pi (3)(2) = 6\pi$$.

Which trigonometric substitution should you choose for radical expressions?

Selecting the correct trigonometric substitution depends on the algebraic structure inside the square root:

What are the most common pitfalls when solving this integral?

• Forgetting to convert dx: You might replace $$dx$$ with $$d\theta$$ instead of $$\cos\theta\,d\theta$$. This drops a factor of $$\cos\theta$$. As a result, it leads to the incorrect integral $$\int \cos\theta\,d\theta$$ instead of $$\int \cos^2\theta\,d\theta$$.
• Leaving the final answer in terms of θ: You must convert an indefinite integral in $$x$$ back to $$x$$. Use $$\theta = \arcsin\left(\frac{x}{a}\right)$$. Also, use a reference right triangle.
• Mixing up limits during substitution: If you change integration bounds to $$\theta$$, do not plug $$x$$-values back in. Conversely, if you convert the antiderivative back to $$x$$, keep the original $$x$$ bounds.
• Confusing half-disks and quarter-disks: Integrating from $$-a$$ to $$a$$ covers a half-disk. This equals $$\frac{\pi a^2}{2}$$. However, integrating from $$0$$ to $$a$$ covers a quarter-disk. This equals $$\frac{\pi a^2}{4}$$.
• Confusing the root integral with its reciprocal: The integral $$\int \frac{dx}{\sqrt{1-x^2}}$$ is a simple inverse sine form. It equals $$\arcsin x + C$$. However, $$\int \sqrt{1-x^2}\,dx$$ requires full trigonometric substitution or integration by parts.

Where is this integral applied in calculus and physics?

The integral of $$\sqrt{1-x^2}$$ serves as a foundational building block across several higher-level topics:

• Area proofs: This method provides a rigorous calculus proof. Also, it validates the geometric area formulas of circles ($$\pi r^2$$) and ellipses ($$\pi a b$$).
• Area between curves: Used when computing area between curves where one or both bounding functions are circular arcs.
• Arc length calculations: Appears in calculus problems involving path lengths and surface areas of revolution for spherical shapes.
• Probability theory: Defines the probability density function of Wigner’s semicircle distribution in random matrix theory.

Evaluate it with the solver

For step-by-step working on other intervals or radii, try the definite integral calculator or, for the antiderivative, the integral calculator.

Practice problems and solutions

Test your understanding with these practice problems before checking the step-by-step solutions below.

1. $$\int_0^2 \sqrt{4 - x^2}\,dx$$
2. $$\int x\sqrt{1 - x^2}\,dx$$
3. $$\int_{-1}^{1} 2\sqrt{1-x^2}\,dx$$
4. $$\int_{-2}^{2} \sqrt{4 - x^2}\,dx$$
5. $$\int \sqrt{9 - x^2}\,dx$$
6. $$\int_0^4 \sqrt{8x - x^2}\,dx$$
7. $$4\int_0^5 \sqrt{25 - x^2}\,dx$$

Answers and step-by-step solutions:

1. $$\pi$$: This represents a quarter-disk of radius $$a = 2$$. So, Area = $$\frac{\pi(2)^2}{4} = \pi$$.
2. $$-\frac{1}{3}(1 - x^2)^{3/2} + C$$: Use standard u-substitution with $$u = 1 - x^2$$ and $$du = -2x,dx$$, not trig substitution.
3. $$\pi$$: Factor out 2: $$2 \int_{-1}^1 \sqrt{1-x^2}\,dx = 2\left(\frac{\pi}{2}\right) = \pi$$.
4. $$2\pi$$: Semicircle of radius $$a = 2$$. Area = $$\frac{\pi(2)^2}{2} = 2\pi$$.
5. $$\frac{x\sqrt{9-x^2}}{2} + \frac{9}{2}\arcsin\left(\frac{x}{3}\right) + C$$: General radius formula with $$a = 3$$.
6. $$4\pi$$: Complete the square: $$8x - x^2 = 16 - (x-4)^2$$. With $$u = x-4$$, bounds $$[0,4]$$ convert to $$[-4,0]$$, representing a quarter-disk of radius 4. As a result, Area = $$\frac{\pi(4)^2}{4} = 4\pi$$.
7. $$25\pi$$: Four times a quarter-disk of radius 5 equals the full disk area: $$4 \cdot \frac{\pi(5)^2}{4} = 25\pi$$.

Frequently asked questions

Can u-substitution solve this integral?

No, standard u-substitution cannot solve $$\int \sqrt{1-x^2}\,dx$$ directly. This is because there is no outer $$x$$ factor. Therefore, it cannot match $$du = -2x,dx$$. You must use trigonometric substitution or integration by parts instead. However, if an $$x$$ factor is present, u-substitution works. For example, in $$\int x\sqrt{1-x^2}\,dx$$, standard u-substitution yields $$-\frac{1}{3}(1-x^2)^{3/2} + C$$.

What is the integral of sqrt(1-x²) from 0 to 1?

The integral $$\int_0^1 \sqrt{1-x^2}\,dx$$ equals $$\frac{\pi}{4}$$. Geometrically, this represents the area of a quarter of the unit circle in the first quadrant.

Why does arcsin appear in the answer?

The arcsine term appears because undoing the trigonometric substitution $$x = \sin\theta$$ requires taking the inverse sine, $$\theta = \arcsin x$$. Geometrically, this arcsine term accounts for the area of the circular sector under the curve.

How do I integrate sqrt(a² – x²)?

Substitute $$x = a\sin\theta$$ with $$dx = a\cos\theta\,d\theta$$. So, the antiderivative is $$\frac{x\sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\arcsin\left(\frac{x}{a}\right) + C$$.

What is the derivative of sqrt(1-x²)?

Applying the chain rule gives $$\frac{d}{dx}\sqrt{1-x^2} = \frac{-x}{\sqrt{1-x^2}}$$. Differentiating the square root is a single-step chain rule problem, whereas integrating it requires trigonometric substitution.

Could I use x = cos θ instead?

Yes, set $$x = \cos\theta$$ for $$0 \le \theta \le \pi$$. This produces $$dx = -\sin\theta\,d\theta$$. As a result, the integral transforms into $$-\int \sin^2\theta\,d\theta$$. The antiderivative then contains $$-\arccos x$$ instead of $$\arcsin x$$. This is mathematically equivalent. Remember, $$\arcsin x + \arccos x = \frac{\pi}{2}$$.

Further reading

• Trig Substitutions (Paul’s Online Math Notes) — worked examples of all three substitutions, including completing the square first.
• 3.3 Trigonometric Substitution (OpenStax Calculus Volume 2) — a textbook section that works through $$\int \sqrt{9 - x^2}\,dx$$ and more.

Related: derivative of arcsin x, integration by parts.