Why These Antiderivatives Are Direct
Some trigonometric antiderivatives are easy to recognize because their integrands are derivatives of familiar functions. The derivative of tangent gives a direct route to an antiderivative of secant squared. The derivative of cotangent gives a route to an antiderivative of cosecant squared, with an important negative sign.
The sign matters: an antiderivative is checked by differentiation, and differentiating cotangent produces negative cosecant squared. The chain rule matters too when the angle is not just \(x\). In this tutorial, we use those derivative facts to obtain the formulas, apply them to several integrals, and check every result by differentiating.
The Basic Derivative Facts
The trigonometric derivative identities used here are \(\frac{d}{dx}\tan x=\sec^2 x\) wherever tangent is defined, and \(\frac{d}{dx}\cot x=-\csc^2 x\) wherever cotangent is defined. Here \(\sec x=1/\cos x\) and \(\csc x=1/\sin x\). Thus secant squared is the derivative of tangent, while the derivative of cotangent is the negative of cosecant squared.
These functions are not defined everywhere. Tangent and secant squared require \(\cos x\ne 0\); cotangent and cosecant squared require \(\sin x\ne 0\). The formulas below apply on any interval where the integrand and the proposed antiderivative are defined.
The Two Basic Formulas
To find an antiderivative of \(\sec^2 x\), recognize it as the derivative of \(\tan x\). To find an antiderivative of \(\csc^2 x\), use the cotangent derivative and reverse its sign.
The constant \(C\) is included because the derivative of a constant is zero. Therefore, adding any constant to an antiderivative does not change its derivative. On an interval, any two antiderivatives of the same function differ by a constant: their difference has derivative zero, and the Mean Value Theorem implies that a differentiable function with derivative zero throughout an interval is constant there.
The second formula is worth checking directly: \(\frac{d}{dx}(-\cot x)=-(-\csc^2 x)=\csc^2 x\). Without the negative sign in front of cotangent, differentiation would give the wrong sign.
Accounting for a Linear Argument
When the trigonometric function has an argument \(ax+b\), where \(a\ne 0\), the chain rule contributes a factor of \(a\) when the proposed antiderivative is differentiated. The antiderivative must therefore include a factor of \(1/a\) to cancel it. For a constant multiplier \(k\), the resulting formulas are:
These formulas require \(a\ne 0\) and are used on intervals where the trigonometric functions involved are defined. For example, the secant-squared formula requires \(\cos(ax+b)\ne 0\) throughout the interval; the cosecant-squared formula requires \(\sin(ax+b)\ne 0\). If \(a=0\), the argument is constant, so the displayed formulas involving division by \(a\) do not apply.
Worked Examples
Worked Example: The Antiderivative of a Basic Secant-Squared Function
Find \(\int 3\sec^2 x\,dx\), and check the result.
The factor \(3\) is constant, and \(\tan x\) differentiates to \(\sec^2 x\). Thus:
To check, differentiate the proposed antiderivative:
This matches the integrand. The formula applies on any interval where \(\cos x\ne 0\).
Worked Example: A Secant-Squared Function with a Linear Argument
Find \(\int 5\sec^2(3x-\frac{\pi}{6})\,dx\), and check the result.
The inside function is \(3x-\frac{\pi}{6}\), whose derivative is \(3\). Applying the linear-argument formula with \(k=5\) and \(a=3\) gives:
Differentiate the proposed antiderivative. The derivative of the outside tangent contributes the secant squared, and the derivative of the inside contributes \(3\):
The result differentiates to the original integrand. The formula is valid on intervals where \(\cos(3x-\frac{\pi}{6})\ne 0\).
Worked Example: A Cosecant-Squared Function with a Linear Argument
Find \(\int 2\csc^2(5x+\frac{\pi}{4})\,dx\), and check the result.
The derivative of the inside function \(5x+\frac{\pi}{4}\) is \(5\). Because the antiderivative of \(\csc^2 u\) is \(-\cot u\), the formula includes both a negative sign and a factor of \(1/5\):
Differentiate to check the sign and coefficient:
The result matches the integrand. It is valid on intervals where \(\sin(5x+\frac{\pi}{4})\ne 0\).
Worked Example: A Negative Cosecant-Squared Integrand
Find \(\int -4\csc^2(2x)\,dx\), and check the result.
Here \(k=-4\) and \(a=2\). The formula gives \(-k/a=-(-4)/2=2\), so:
Differentiate:
The check recovers the negative coefficient in the original integrand. This example shows why the signs in the formula should be handled as algebra, rather than guessed.
Why the Formulas Work
The linear-argument formulas follow directly from the chain rule and the basic derivative identities. Here is the verification for both formulas. Suppose \(a\ne 0\), and consider intervals where the functions are defined.
For \(F(x)=\frac{k}{a}\tan(ax+b)\), the chain rule gives \(F'(x)=\frac{k}{a}\sec^2(ax+b)\cdot a=k\sec^2(ax+b)\). Thus \(F\) has the required derivative.
For \(G(x)=-\frac{k}{a}\cot(ax+b)\), the chain rule gives \(G'(x)=-\frac{k}{a}\bigl[-\csc^2(ax+b)\cdot a\bigr]=k\csc^2(ax+b)\). The two negative signs cancel.
These calculations establish the formulas: each proposed function differentiates to the integrand. They also show exactly where the coefficient \(1/a\) comes from. The inner derivative contributes \(a\), so the coefficient outside the trigonometric function must cancel that factor.
Common Mistake: Losing the Negative Sign
A frequent error is to write \(\int \csc^2 x\,dx=\cot x+C\). This fails when checked: \(\frac{d}{dx}(\cot x+C)=-\csc^2 x\), which is the negative of the desired integrand. The correction is:
Another error is to apply the chain-rule factor twice. For example, differentiating \(\tan(4x)\) gives \(\sec^2(4x)\cdot 4\), not \(4\sec^2(4x)\cdot 4\). The factor \(4\) appears once because it is the derivative of the inside function. In an antiderivative, the corresponding coefficient \(1/4\) cancels that single factor.
Choosing and Checking the Formula
Before writing an antiderivative, identify which derivative identity matches the integrand. Then keep track of any constant multiplier and the derivative of the angle. A quick differentiation check is especially useful for confirming the cosecant-squared sign.
- For \(\sec^2\) with argument \(ax+b\), use tangent and include \(1/a\).
- For \(\csc^2\) with argument \(ax+b\), use negative cotangent and include \(1/a\).
- Multiply the result by any constant already present in the integrand.
- Differentiate the proposed answer, applying the chain rule exactly once to the inside function.
- Check that the integrand and the antiderivative are defined on the interval in question.
These checks turn two short formulas into reliable tools. In each case, the proposed antiderivative is not accepted merely because it resembles a memorized pattern: its derivative must reproduce the original function, including its sign and every numerical factor.
Check Your Understanding
For each question, pay attention to the sign, the coefficient, and any restriction on where the functions are defined.
- What is \(\int \sec^2 x\,dx\), and how can differentiation verify it?
- Why does the antiderivative of \(\csc^2 x\) require a negative cotangent?
- Find an antiderivative of \(7\sec^2(2x+1)\).
- Find an antiderivative of \(3\csc^2(4x-\frac{\pi}{3})\).
- What condition on \(a\) is needed to use the displayed linear-argument formulas, and what factor does the chain rule produce?
