The indefinite integral of sec²(x) with respect to x is tan(x) + C. Here, C is the constant of integration. The derivative of tan(x) equals sec²(x). Therefore, you can calculate the integral of sec²(x). This directly reverses differentiation. As a result, it yields tan(x) + C.
The integral of sec²(x) is a short, fundamental calculus result. However, it is also a vital building block. You will use it for more complex integrals. These include tan²(x) and higher even powers of secant. You also use it for trigonometric substitution and rational trig expressions. Understanding this antiderivative makes solving harder multi-step problems significantly easier.
Why is the integral of sec²(x) equal to tan(x) + C?
Integration is the inverse operation of differentiation. To find the integral of sec²(x), you look for a function whose derivative is sec²(x). That function is tan(x).
To prove this directly, write tangent as the quotient of sine and cosine. Then, evaluate the derivative of tan x. Use the quotient rule for this:
$$\begin{aligned}\frac{d}{dx} \left(\frac{\sin x}{\cos x}\right) &= \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} &= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} &= \frac{1}{\cos^2 x} = \sec^2 x\end{aligned}$$
The Pythagorean identity \(\cos^2 x + \sin^2 x = 1\) simplifies the numerator to 1. Reading this derivative equation in reverse gives the integral formula \(\int \sec^2 x \, dx = \tan x + C\). Memorizing both directions together prevents sign confusion during exam problems.
What does sec²(x) represent geometrically?
Because \(\sec^2 x = \frac{1}{\cos^2 x}\), \(\sec^2 x\) is always greater than or equal to 1. Also, \(\cos^2 x \le 1\) for all real numbers. This applies where cosine is defined. Geometrically, \(\sec^2 x\) represents the instantaneous slope of \(y = \tan x\). The graph of \(\tan x\) increases continuously. This occurs on every open interval between its vertical asymptotes. It has a minimum slope of 1 at \(x = 0\). However, the slope grows infinitely steep as \(x\) approaches \(\frac{\pi}{2}\).
According to the Fundamental Theorem of Calculus, you can find the area under the curve \(y = \sec^2 x\). This area is over an interval $[0, b]$. As a result, it equals the net change in height of the tangent function across that same interval:
$$\int_0^b \sec^2 x \, dx = \tan b - \tan 0 = \tan b$$
Evaluating this definite integral on the interval \(\left[0, \frac{\pi}{4}\right]\) illustrates the relationship:
$$\int_0^{\pi/4} \sec^2 x \, dx = \tan\left(\frac{\pi}{4}\right) - \tan(0) = 1 - 0 = 1$$
On the interval \(\left[0, \frac{\pi}{4}\right]\), the function \(\sec^2 x\) increases smoothly from 1 to 2. The curve’s area equals 1. This occurs over an interval width of approximately 0.785. As a result, this matches geometric expectations.
How do you integrate sec²(x) with inside functions or u-substitution?
When the argument inside secant-squared is a linear function \(ax + b\), you must apply the chain rule in reverse. This requires you to divide by the coefficient \(a\).
$$\int \sec^2(ax + b) \, dx = \frac{\tan(ax + b)}{a} + C$$
For example, integrating \(\sec^2(3x)\) gives:
$$\int \sec^2(3x) \, dx = \frac{\tan(3x)}{3} + C$$
Differentiating \(\tan(3x)\) produces an extra factor of 3. This happens because of the chain rule. So, you can divide by 3 to cancel it. If the inside function is non-linear, you can evaluate the integral. Use u-substitution for this. However, the inner function’s derivative must be present as an outside multiplier.
$$\int x \sec^2(x^2) \, dx$$
Let \(u = x^2\). This means \(du = 2x \, dx\). You can also write this as \(\frac{1}{2} du = x \, dx\). Substituting these terms simplifies the integral.
$$\int x \sec^2(x^2) \, dx = \frac{1}{2} \int \sec^2 u \, du = \frac{1}{2} \tan u + C = \frac{\tan(x^2)}{2} + C$$
How do you integrate tan²(x)?
A common mistake in calculus is trying to apply the power rule to \(\int \tan^2 x \, dx\). Instead, rewrite the integrand using the trigonometric identity \(\tan^2 x = \sec^2 x - 1\):
$$\int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \int \sec^2 x \, dx - \int 1 \, dx = \tan x - x + C$$
Converting \(\tan^2 x\) into secant-squared turns an otherwise difficult expression into a single-step basic antiderivative.
How do you spot sec²(x) hidden in disguise?
You can simplify many trigonometric integrals. First, rewrite the integrand. This isolates \(\sec^2 x\). Also, look for these common algebraic transformations:
- \(\frac{1}{\cos^2 x}\), which is equivalent to \(\sec^2 x\) by definition.
- \(1 + \tan^2 x\), which equals \(\sec^2 x\) via the Pythagorean identity.
- Also consider \(\frac{1}{1 + \cos 2x}\) or \(\frac{1}{1 - \sin^2 x}\). These simplify directly to constant multiples of \(\frac{1}{\cos^2 x}\).
- Products where \(\sec^2 x\) multiplies a power or function of \(\tan x\). Here, \(\sec^2 x \, dx\) serves as \(du\) for the substitution \(u = \tan x\).
Whenever \(\tan x\) appears inside a complex expression, \(\sec^2 x\) may also be present. You can set \(u = \tan x\). This converts the entire integrand into an algebraic polynomial or rational function.
Step-by-step worked examples
Example 1 (Definite interval evaluation). Calculate the definite integral of \(\sec^2 x\) from \(x = \frac{\pi}{6}\) to \(x = \frac{\pi}{3}\):
$$\int_{\pi/6}^{\pi/3} \sec^2 x \, dx = \left[ \tan x \right]_{\pi/6}^{\pi/3} = \tan\left(\frac{\pi}{3}\right) - \tan\left(\frac{\pi}{6}\right) = \sqrt{3} - \frac{\sqrt{3}}{3} = \frac{2\sqrt{3}}{3} \approx 1.1547$$
Example 2 (Trigonometric identity substitution). Evaluate \(\int \frac{dx}{1 + \cos 2x}\). Use the double-angle identity \(1 + \cos 2x = 2\cos^2 x\). Then, substitute this into the denominator.
$$\int \frac{dx}{1 + \cos 2x} = \int \frac{dx}{2\cos^2 x} = \frac{1}{2} \int \sec^2 x \, dx = \frac{\tan x}{2} + C$$
You can instantly simplify rational trigonometric integrals. First, recognize that a cosine-squared term in the denominator forms a secant-squared term.
Example 3 (Power of tangent with secant-squared). Evaluate \(\int \sec^2 x \tan^3 x \, dx\). First, let \(u = \tan x\). As a result, \(du = \sec^2 x \, dx\).
$$\int \sec^2 x \tan^3 x \, dx = \int u^3 \, du = \frac{u^4}{4} + C = \frac{\tan^4 x}{4} + C$$
Example 4 (Logarithmic antiderivative form). Evaluate \(\int \frac{\sec^2 x}{1 + \tan x} \, dx\). The numerator is the derivative of the denominator. Let \(u = 1 + \tan x\), so \(du\) is the integral of sec^2x, \(dx\):
$$\int \frac{\sec^2 x}{1 + \tan x} \, dx = \int \frac{du}{u} = \ln|u| + C = \ln|1 + \tan x| + C$$
Keep absolute value bars around the logarithm argument because \(1 + \tan x\) takes negative values whenever \(\tan x < -1\).
Integrating sec²(x) e^(tan x)
Example 5 (Exponential function with tangent argument). Evaluate \(\int \sec^2 x , e^{\tan x} \, dx\). You can use \(u = \tan x\). Then \(du = \sec^2 x \, dx\).
$$\int \sec^2 x , e^{\tan x} \, dx = \int e^u \, du = e^u + C = e^{\tan x} + C$$
You can confirm this result by differentiating \(e^{\tan x}\). Use the chain rule. Because of this, \(\frac{d}{dx}\left(e^{\tan x}\right) = e^{\tan x} \cdot \sec^2 x\).
Example 6 (Integration by parts). You can evaluate \(\int x \sec^2 x \, dx\). First, set \(u = x\). Also set \(dv = \sec^2 x \, dx\). This yields \(du = dx\) and \(v = \tan x\). Next, apply the integration by parts formula: \(\int u \, dv = uv - \int v \, du\). This gives you:
$$\int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx$$
The remaining term requires the integral of tan x. It equals \(-\ln|\cos x| + C\). You can also write it as \(\ln|\sec x| + C\).
$$\int x \sec^2 x \, dx = x \tan x + \ln|\cos x| + C$$
Example 7 (Binomial square expansion). Evaluate \(\int (\sec x + \tan x)^2 \, dx\). Expanding the square gives \(\sec^2 x + 2\sec x \tan x + \tan^2 x\). Next, replacing \(\tan^2 x\) with \(\sec^2 x - 1\) combines like terms:
$$\int (\sec x + \tan x)^2 \, dx = \int (2\sec^2 x + 2\sec x \tan x - 1) \, dx = 2\tan x + 2\sec x - x + C$$
You can check worked solutions using an integral calculator.
How does sec²(x) compare to other basic trig integrals?
The table below compares the antiderivative of \(\sec^2 x\) to other standard trigonometric integrals taught in calculus:
| Integral | Result |
|---|---|
| \(\int \sec^2 x \, dx\) | \(\tan x + C\) |
| \(\int \csc^2 x \, dx\) | \(-\cot x + C\) |
| \(\int \sec x \tan x \, dx\) | \(\sec x + C\) |
| \(\int \csc x \cot x \, dx\) | \(-\csc x + C\) |
| \(\int \sec x \, dx\) | \(\ln|\sec x + \tan x| + C\) |
Integrating \(\sec x\) requires a specific algebraic multiplication trick; see the full breakdown for the integral of sec x.
The csc²(x) twin integral
Cosecant-squared follows the same pattern as secant-squared, but incorporates a negative sign. Because the derivative of \(\cot x\) is \(-\csc^2 x\), reversing the operation yields:
$$\int \csc^2 x \, dx = -\cot x + C$$
A reliable memory rule: every cofunction derivative in calculus (cosine, cotangent, cosecant) produces a negative sign.
What are common mistakes when integrating sec²(x)?
- Incorrectly applying the power rule. Writing \(\int \sec^2 x \, dx = \frac{\sec^3 x}{3}\) is wrong. Differentiating \(\frac{\sec^3 x}{3}\) produces \(\sec^3 x \tan x\), not \(\sec^2 x\).
- Confusing sec²(x) with sec(x). Plain \(\sec x\) integrates to \(\ln|\sec x + \tan x| + C\), whereas \(\sec^2 x\) integrates directly to \(\tan x + C\).
- Forgetting inner linear scaling factors. The integral \(\int \sec^2(4x) \, dx\) is \(\frac{\tan(4x)}{4} + C\), not \(\tan(4x) + C\).
- Integrating across vertical asymptotes. Evaluating \(\int_0^{\pi} \sec^2 x \, dx\) by calculating \(\tan \pi - \tan 0 = 0\) is invalid. The function \(\sec^2 x\) has an asymptote at \(x = \frac{\pi}{2}\) where it goes to infinity, so the integral diverges.
- Attempting power rule integration on tan²(x). You cannot evaluate \(\int \tan^2 x \, dx\) as \(\frac{\tan^3 x}{3}\) because the derivative of tangent (\(\sec^2 x\)) is missing. Rewrite \(\tan^2 x\) as \(\sec^2 x - 1\) first.
- Confusing sec²(x) with sec(x²). The expression \(\sec^2 x\) means \((\sec x)^2\). However, \(\sec(x^2)\) squares the input argument \(x\). The formula \(\tan x + C\) applies only to \(\sec^2 x\).
Where is the integral of sec²(x) used in calculus?
The integral of sec^2x is fundamental for trigonometric substitution. Use the substitution \(x = \tan \theta\) when solving integrals. These integrals contain \(\sqrt{1 + x^2}\) or \(1 + x^2\). This sets \(dx = \sec^2 \theta \, d\theta\). Also, \(1 + x^2 = \sec^2 \theta\). For example, evaluating \(\int \frac{dx}{1 + x^2}\) yields:
$$\int \frac{\sec^2 \theta \, d\theta}{\sec^2 \theta} = \int 1 \, d\theta = \theta + C = \arctan x + C$$
It is also essential for integrating higher even powers of secant. For example, consider \(\sec^4 x\) or \(\sec^6 x\). For \(\int \sec^4 x \, dx\), separate one factor of \(\sec^2 x\). This will serve as \(du\). Then, rewrite the remaining \(\sec^2 x\) as \(1 + \tan^2 x\).
$$\int \sec^4 x \, dx = \int (1 + \tan^2 x) \sec^2 x \, dx$$
Let \(u = \tan x\) and \(du = \sec^2 x \, dx\):
$$\int (1 + u^2) \, du = u + \frac{u^3}{3} + C = \tan x + \frac{\tan^3 x}{3} + C$$
Practice problems with step-by-step solutions
- \(\int \sec^2(4x) \, dx\)
- \(\int_0^{\pi/3} \sec^2 x \, dx\)
- \(\int (1 + \tan^2 x) \, dx\)
- \(\int \sec^2\left(\frac{x}{2}\right) \, dx\)
- \(\int_0^{\pi/12} \sec^2(3x) \, dx\)
- \(\int \sec^2 x \tan x \, dx\)
Solutions:
- Using linear substitution \(u = 4x\): \(\frac{\tan(4x)}{4} + C\).
- Evaluating FTC limits: \(\tan\left(\frac{\pi}{3}\right) - \tan(0) = \sqrt{3} - 0 = \sqrt{3}\).
- Applying the identity \(1 + \tan^2 x = \sec^2 x\): \(\int \sec^2 x \, dx = \tan x + C\).
- Dividing by the linear factor \(\frac{1}{2}\): \(2\tan\left(\frac{x}{2}\right) + C\).
- Evaluating FTC limits: \(\left[ \frac{\tan(3x)}{3} \right]_0^{\pi/12} = \frac{\tan(\pi/4)}{3} - \frac{\tan(0)}{3} = \frac{1}{3} - 0 = \frac{1}{3}\).
- Let \(u = \tan x\). Then \(du = \sec^2 x \, dx\). As a result, \(\int u \, du = \frac{u^2}{2} + C = \frac{\tan^2 x}{2} + C\). Alternatively, using \(u = \sec x\) yields \(\frac{\sec^2 x}{2} + C\). This differs only by a constant. Because \(\frac{\sec^2 x}{2} = \frac{\tan^2 x + 1}{2} = \frac{\tan^2 x}{2} + \frac{1}{2}\).
Frequently asked questions
Is the integral of sec²(x) equal to sec³(x)/3?
No, the power rule does not apply to \(\sec^2 x\) alone. This is because the derivative of secant, which is \(\sec x \tan x\), is missing. Differentiating \(\frac{\sec^3 x}{3}\) yields \(\sec^3 x \tan x\). It does not yield the integral of sec^2x.
What is the integral of sec²(x) tan(x)?
Setting \(u = \tan x\) gives \(du = \sec^2 x \, dx\), turning the integral into \(\int u \, du = \frac{\tan^2 x}{2} + C\). However, setting \(u = \sec x\) yields \(\frac{\sec^2 x}{2} + C\).
What is the integral of sec²(x) from 0 to π/2?
The definite integral diverges to infinity. As \(x\) approaches \(\frac{\pi}{2}\) from the left, \(\tan x\) grows without bound. So, the area under the integral of sec^2x is infinite.
Why is there no logarithm in the integral of sec²(x)?
Logarithms appear in antiderivatives when integrating rational functions of the form \(\frac{f'(x)}{f(x)}\). Because \(\sec^2 x\) is already the direct derivative of \(\tan x\), its antiderivative is simply \(\tan x + C\).
Further reading
- Derivatives of Trig Functions (Paul’s Online Math Notes) — detailed derivations showing how the quotient rule establishes trigonometric derivatives.
- Integrals Involving Trig Functions (Paul’s Online Math Notes) — strategies for integrating powers of secant and tangent.
- Computing Indefinite Integrals (Paul’s Online Math Notes) — comprehensive tables and techniques for fundamental antiderivatives.
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