Q. \[ \int \sqrt{1-x^2}\,dx \]

Q: How do you solve it using the trig substitution x=\sin\theta?

Let x=\sin\theta. Then \sqrt{1-x^2}=\cos\theta and dx=\cos\theta\,d\theta. Integral becomes \int \cos^2\theta\,d\theta=\frac{1}{2}\left(\theta+\sin\theta\cos\theta\right)+C. Convert back: \theta=\arcsin(x), \sin\theta=x, \cos\theta=\sqrt{1-x^2}.

Q: How do you check the derivative of \frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\arcsin(x) ?

Differentiate term-by-term. The derivative of \frac{x}{2}\sqrt{1-x^2} simplifies to \frac{1-2x^2}{2\sqrt{1-x^2}}, and \frac{d}{dx}\left(\frac{1}{2}\arcsin(x)\right)=\frac{1}{2\sqrt{1-x^2}}. Sum gives \sqrt{1-x^2}.

Q: What is \int_{-1}^{1} \sqrt{1-x^2}\,dx ?

This is the area of a semicircle of radius 1: \[ \int_{-1}^{1} \sqrt{1-x^2}\,dx=\frac{\pi}{2}. \]

Q: Can we express the result using \arccos(x) instead of \arcsin(x) ?

Using \arcsin(x)+\arccos(x)=\frac{\pi}{2} , you can rewrite \[ \frac{1}{2}\arcsin(x)= -\frac{1}{2}\arccos(x)+\frac{\pi}{4}. \] So the integral is \[ \frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}\arccos(x)+C. \]

Answer

We compute

\[
\int \sqrt{1-x^2}\,dx.
\]

Use the trig substitution \(x=\sin\theta\). Then \(dx=\cos\theta\,d\theta\) and \(\sqrt{1-x^2}=\cos\theta\). So

\[
\int \sqrt{1-x^2}\,dx=\int \cos\theta\cdot \cos\theta\,d\theta=\int \cos^2\theta\,d\theta.
\]

Apply \(\cos^2\theta=\frac{1+\cos 2\theta}{2}\):

\[
\int \cos^2\theta\,d\theta=\frac{1}{2}\int (1+\cos 2\theta)\,d\theta=\frac{\theta}{2}+\frac{\sin 2\theta}{4}+C.
\]

Convert back. Since \(x=\sin\theta\), we have \(\theta=\arcsin x\) and

\[
\sin 2\theta=2\sin\theta\cos\theta=2x\sqrt{1-x^2}.
\]

Therefore the integral is

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

Detailed Explanation

We want to compute the indefinite integral:

\[
\int \sqrt{1-x^2}\,dx
\]

This integrand suggests using a trigonometric substitution, because \(1-x^2\) resembles the Pythagorean identity.

Step 1: Choose a trigonometric substitution

Let

\[
x=\sin\theta
\]

Then \(dx=\cos\theta\,d\theta\).

Also, compute the expression under the square root:

\[
1-x^2=1-\sin^2\theta=\cos^2\theta
\]

\[
\sqrt{1-x^2}=\sqrt{\cos^2\theta}=|\cos\theta|
\]

For a standard antiderivative, we take \(\sqrt{\cos^2\theta}=\cos\theta\) (the final result can be written in a form that works for the typical domain of the antiderivative). Thus:

\[
\sqrt{1-x^2}=\cos\theta
\]

Step 2: Substitute into the integral

Substitute \(x=\sin\theta\), \(\sqrt{1-x^2}=\cos\theta\), and \(dx=\cos\theta\,d\theta\):

\[
\int \sqrt{1-x^2}\,dx=\int \cos\theta\cdot \cos\theta\,d\theta=\int \cos^2\theta\,d\theta
\]

Step 3: Use the power-reduction identity

Recall the identity:

\[
\cos^2\theta=\frac{1+\cos(2\theta)}{2}
\]

So the integral becomes:

\[
\int \cos^2\theta\,d\theta=\int \frac{1+\cos(2\theta)}{2}\,d\theta
\]

Distribute the \(\frac{1}{2}\):

\[
\int \frac{1+\cos(2\theta)}{2}\,d\theta=\frac{1}{2}\int 1\,d\theta+\frac{1}{2}\int \cos(2\theta)\,d\theta
\]

Step 4: Integrate term by term

\[
\frac{1}{2}\int 1\,d\theta=\frac{1}{2}\theta
\]

For the second integral, use:

\[
\int \cos(2\theta)\,d\theta=\frac{1}{2}\sin(2\theta)
\]

Therefore:

\[
\frac{1}{2}\int \cos(2\theta)\,d\theta=\frac{1}{2}\cdot \frac{1}{2}\sin(2\theta)=\frac{1}{4}\sin(2\theta)
\]

So the integral in \(\theta\) is:

\[
\int \cos^2\theta\,d\theta=\frac{1}{2}\theta+\frac{1}{4}\sin(2\theta)+C
\]

Step 5: Rewrite everything back in terms of \(x\)

We know \(x=\sin\theta\), so:

\[
\theta=\arcsin(x)
\]

Next compute \(\sin(2\theta)\). Use the identity:

\[
\sin(2\theta)=2\sin\theta\cos\theta
\]

Since \(\sin\theta=x\) and \(\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-x^2}\), we get:

\[
\sin(2\theta)=2x\sqrt{1-x^2}
\]

Substitute both expressions into the result:

\[
\int \sqrt{1-x^2}\,dx=\frac{1}{2}\arcsin(x)+\frac{1}{4}\left(2x\sqrt{1-x^2}\right)+C
\]

Simplify the coefficient:

\[
\frac{1}{4}\cdot 2x\sqrt{1-x^2}=\frac{1}{2}x\sqrt{1-x^2}
\]

Final Answer

\[
\int \sqrt{1-x^2}\,dx=\frac{1}{2}\arcsin(x)+\frac{1}{2}x\sqrt{1-x^2}+C
\]

This is a correct antiderivative.

See full solution

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Calculus FAQ

What is \( \int \sqrt{1-x^2}\,dx \)?

\[ \int \sqrt{1-x^2}\,dx=\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\arcsin(x)+C. \]

How do you solve it using the trig substitution \(x=\sin\theta\)?

Let \(x=\sin\theta\). Then \(\sqrt{1-x^2}=\cos\theta\) and \(dx=\cos\theta\,d\theta\). Integral becomes \(\int \cos^2\theta\,d\theta=\frac{1}{2}\left(\theta+\sin\theta\cos\theta\right)+C\). Convert back: \(\theta=\arcsin(x)\), \(\sin\theta=x\), \(\cos\theta=\sqrt{1-x^2}\).

What is \( \int_0^1 \sqrt{1-x^2}\,dx \)?

This equals the area of a quarter unit circle: \[ \int_0^1 \sqrt{1-x^2}\,dx=\frac{\pi}{4}. \]

How do you check the derivative of \( \frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\arcsin(x) \)?

Differentiate term-by-term. The derivative of \(\frac{x}{2}\sqrt{1-x^2}\) simplifies to \(\frac{1-2x^2}{2\sqrt{1-x^2}}\), and \(\frac{d}{dx}\left(\frac{1}{2}\arcsin(x)\right)=\frac{1}{2\sqrt{1-x^2}}\). Sum gives \(\sqrt{1-x^2}\).

What is \( \int_{-1}^{1} \sqrt{1-x^2}\,dx \)?

This is the area of a semicircle of radius \(1\): \[ \int_{-1}^{1} \sqrt{1-x^2}\,dx=\frac{\pi}{2}. \]

Can we express the result using \( \arccos(x) \) instead of \( \arcsin(x) \)?

Using \( \arcsin(x)+\arccos(x)=\frac{\pi}{2} \), you can rewrite \[ \frac{1}{2}\arcsin(x)= -\frac{1}{2}\arccos(x)+\frac{\pi}{4}. \] So the integral is \[ \frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}\arccos(x)+C. \]

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