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

Q: . How do I compute \int \frac{1}{\sqrt{1-x^2}}\,dx ?

. Use x=\sin\theta. Then dx=\cos\theta\,d\theta and \sqrt{1-x^2}=\cos\theta. Integral becomes \int d\theta=\theta+C=\arcsin(x)+C.

Q: . What’s the equivalent form using inverse cosine?

. Since \arcsin(x)+\arccos(x)=\frac{\pi}{2} , both differ by a constant. Thus \int \frac{1}{\sqrt{1-x^2}}\,dx = \arccos(x)+C as well.

Q: . What substitution works fastest: x=\sin\theta or x=\cos\theta?

. Either works. With x=\sin\theta, \sqrt{1-x^2}=\cos\theta. With x=\cos\theta, \sqrt{1-x^2}=\sin\theta. Both reduce the integrand to d\theta.

Q: . What domain restrictions apply for the integral?

. For real values, require 1-x^2\ge 0\Rightarrow -1\le x\le 1. The integrand is singular at x=\pm1, so the antiderivative is typically stated for |x|<1.

Q: . How can I verify the answer by differentiating?

. Differentiate \arcsin(x): \frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}} . This matches the integrand, confirming \int \frac{1}{\sqrt{1-x^2}}\,dx=\arcsin(x)+C.

Q: . What does the integral become in trig form, like \theta substitution?

. In the x=\sin\theta substitution, the result is \theta + C. Since \theta=\arcsin(x), the final form is \arcsin(x)+C.

Answer

We need

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

Use the standard derivative

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

So the integral is

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

Detailed Explanation

We want to solve the integral

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

A standard method for integrals of the form \( \frac{1}{\sqrt{1-x^2}} \) is to use a trigonometric substitution. This is because the expression \(1-x^2\) looks like the identity \( \sin^2\theta + \cos^2\theta = 1\).

Step 1: Choose a trigonometric substitution

Let

\[
x=\sin\theta.
\]

Then we have

\[
dx=\cos\theta\,d\theta.
\]

Also, substitute into the square root:

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

Assuming we are in a region where \(\cos\theta \ge 0\), we can take

\[
\sqrt{\cos^2\theta}=\cos\theta.
\]

Step 2: Rewrite the integral in terms of \(\theta\)

Substitute \(x=\sin\theta\) and \(dx=\cos\theta\,d\theta\) into the integral:

\[
\int \frac{1}{\sqrt{1-x^2}}\,dx
=\int \frac{1}{\cos\theta}\,\bigl(\cos\theta\,d\theta\bigr).
\]

Now simplify by canceling \(\cos\theta\):

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

Step 3: Integrate

\[
\int 1\,d\theta = \theta + C,
\]
where \(C\) is the constant of integration.

Step 4: Convert back to \(x\)

We used \(x=\sin\theta\), so \(\theta = \arcsin(x)\). Therefore,

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

Final Answer

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

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Calculus FAQ

. How do I compute \( \int \frac{1}{\sqrt{1-x^2}}\,dx \) ?

. Use \(x=\sin\theta\). Then \(dx=\cos\theta\,d\theta\) and \(\sqrt{1-x^2}=\cos\theta\). Integral becomes \(\int d\theta=\theta+C=\arcsin(x)+C\).

. What’s the equivalent form using inverse cosine?

. Since \( \arcsin(x)+\arccos(x)=\frac{\pi}{2} \), both differ by a constant. Thus \( \int \frac{1}{\sqrt{1-x^2}}\,dx = \arccos(x)+C\) as well.

. What substitution works fastest: \(x=\sin\theta\) or \(x=\cos\theta\)?

. Either works. With \(x=\sin\theta\), \(\sqrt{1-x^2}=\cos\theta\). With \(x=\cos\theta\), \(\sqrt{1-x^2}=\sin\theta\). Both reduce the integrand to \(d\theta\).

. What domain restrictions apply for the integral?

. For real values, require \(1-x^2\ge 0\Rightarrow -1\le x\le 1\). The integrand is singular at \(x=\pm1\), so the antiderivative is typically stated for \(|x|<1\).

. How can I verify the answer by differentiating?

. Differentiate \( \arcsin(x)\): \( \frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}} \). This matches the integrand, confirming \( \int \frac{1}{\sqrt{1-x^2}}\,dx=\arcsin(x)+C\).

. What does the integral become in trig form, like \( \theta \) substitution?

. In the \(x=\sin\theta\) substitution, the result is \( \theta + C\). Since \(\theta=\arcsin(x)\), the final form is \( \arcsin(x)+C\).

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