Q. \[ \int \sec^2(x)\,dx \]

Q: What is the integral \int \sec^{2}(x)\,dx ?

\int \sec^{2}(x)\,dx = \tan(x) + C , since \frac{d}{dx}\tan(x)=\sec^{2}(x) .

Q: How do you prove \frac{d}{dx}\tan(x)=\sec^{2}(x) ?

Let y=\tan(x). Then y'=\sec^{2}(x) by the standard derivative rule \tan'(x)=\sec^{2}(x).

Q: What is \int \sec^{2}(ax)\,dx ?

Use substitution u=ax. Then dx=\frac{1}{a}\,du, so \int \sec^{2}(ax)\,dx=\frac{1}{a}\tan(ax)+C .

Q: What is \int 3\sec^{2}(x)\,dx ?

Pull out the constant: \int 3\sec^{2}(x)\,dx = 3\tan(x)+C .

Q: What is \int \sec^{2}(x)\tan(x)\,dx ? (related via substitution)

Let u=\tan(x), so du=\sec^{2}(x)\,dx. Then \int \sec^{2}(x)\tan(x)\,dx=\int u\,du=\frac{1}{2}\tan^{2}(x)+C .

Q: What is \int \sec^{2}(x)\,d(\tan(x)) ?

Since d(\tan(x))=\sec^{2}(x)\,dx, the integral becomes \int d(\tan(x)) = \tan(x)+C .

Q: Are there restrictions on x for \int \sec^{2}(x)\,dx ?

\sec(x) must be defined, so \cos(x)\neq 0 . The antiderivative \tan(x)+C works on any interval avoiding \cos(x)=0 .

Answer

To integrate \( \sec^2(x) \), use the fact that the derivative of \( \tan(x) \) is \( \sec^2(x) \).

\[
\int \sec^2(x)\,dx=\tan(x)+C
\]

Detailed Explanation

We want to compute the indefinite integral

\[
\int \sec^2(x)\,dx.
\]

Step 1: Recall the key derivative.

Remember that the derivative of \(\tan(x)\) is

\[
\frac{d}{dx}\big(\tan(x)\big)=\sec^2(x).
\]

This is the fundamental identity we will use.

Step 2: Match the integrand to the derivative.

The integrand is exactly \(\sec^2(x)\). Since

\[
\frac{d}{dx}\big(\tan(x)\big)=\sec^2(x),
\]

it means \(\sec^2(x)\) is the derivative of \(\tan(x)\).

Step 3: Integrate by reversing the derivative.

If the derivative of \(\tan(x)\) is \(\sec^2(x)\), then the integral of \(\sec^2(x)\) is \(\tan(x)\) plus a constant.

\[
\int \sec^2(x)\,dx=\tan(x)+C.
\]

Final Answer

\[
\int \sec^2(x)\,dx=\tan(x)+C.
\]

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Calculus FAQ

What is the integral \( \int \sec^{2}(x)\,dx \)?

\( \int \sec^{2}(x)\,dx = \tan(x) + C \), since \( \frac{d}{dx}\tan(x)=\sec^{2}(x) \).

How do you prove \( \frac{d}{dx}\tan(x)=\sec^{2}(x) \)?

Let \(y=\tan(x)\). Then \(y'=\sec^{2}(x)\) by the standard derivative rule \( \tan'(x)=\sec^{2}(x)\).

What is \( \int \sec^{2}(ax)\,dx \)?

Use substitution \(u=ax\). Then \(dx=\frac{1}{a}\,du\), so \( \int \sec^{2}(ax)\,dx=\frac{1}{a}\tan(ax)+C \).

What is \( \int 3\sec^{2}(x)\,dx \)?

Pull out the constant: \( \int 3\sec^{2}(x)\,dx = 3\tan(x)+C \).

What is \( \int \sec^{2}(x)\tan(x)\,dx \)? (related via substitution)

Let \(u=\tan(x)\), so \(du=\sec^{2}(x)\,dx\). Then \( \int \sec^{2}(x)\tan(x)\,dx=\int u\,du=\frac{1}{2}\tan^{2}(x)+C \).

What is \( \int \sec^{2}(x)\,d(\tan(x)) \)?

Since \(d(\tan(x))=\sec^{2}(x)\,dx\), the integral becomes \( \int d(\tan(x)) = \tan(x)+C \).

Are there restrictions on \(x\) for \( \int \sec^{2}(x)\,dx \)?

\(\sec(x)\) must be defined, so \( \cos(x)\neq 0 \). The antiderivative \( \tan(x)+C \) works on any interval avoiding \( \cos(x)=0 \).

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