Why an Odd Power of Tangent Helps
When an integral contains an odd power of tangent and a factor of secant, it contains a useful derivative pattern: the derivative of \(\sec x\) is \(\sec x\tan x\). Reserving one factor of each makes a substitution with \(u=\sec x\) possible. The remaining tangent power is even, so it can be rewritten using \(\tan^2 x=\sec^2 x-1\).
This method is especially systematic because, after the substitution, the integrand is a polynomial in \(u\). The exponent of secant determines the initial power of \(u\), while the even tangent power determines the degree of the additional factor.
The requirement \(n\geq 1\) ensures there is at least one secant factor available to pair with a tangent factor. The substitution is \(u=\sec x\), so \(du=\sec x\tan x\,dx\). The trigonometric identity used to convert the remaining tangent powers follows from \(\sec^2 x=1+\tan^2 x\).
The Substitution, Step by Step
Separate one factor of \(\sec x\tan x\) from the integrand. Since the tangent power is \(2k+1\), the tangent factors left over have the even power \(2k\):
Now replace \(\tan^{2k}(x)\) with \((\sec^2 x-1)^k\). With \(u=\sec x\), the integral becomes a polynomial integral:
Write the integrand as \(\tan^{2k}(x)\sec^{n-1}(x)\bigl(\sec x\tan x\bigr)\,dx\).
Use \(\tan^{2k}(x)=(\tan^2 x)^k=(\sec^2 x-1)^k\).
Set \(u=\sec x\) and \(du=\sec x\tan x\,dx\), then integrate the resulting polynomial.
Replace \(u\) by \(\sec x\), include the constant of integration, and differentiate the result.
Worked Examples
Worked Example: A Cubic Tangent Power and a Secant-Squared Factor
Evaluate \(\int \tan^3 x\sec^2 x\,dx\).
Here \(2k+1=3\), so \(k=1\), and \(n=2\). Reserve \(\sec x\tan x\,dx\). One tangent factor and one secant factor remain:
To check, differentiate the proposed antiderivative:
The derivative is the original integrand, so the antiderivative is verified.
Worked Example: A Fifth-Power Tangent Factor and One Secant
Evaluate \(\int \tan^5 x\sec x\,dx\).
In this case \(k=2\) and \(n=1\). After reserving \(\sec x\tan x\,dx\), four tangent factors remain, and there are no additional secant factors:
Differentiate to check each term:
This recovers the given integrand.
Worked Example: A Cubic Tangent Power and a Cubed Secant Factor
Evaluate \(\int \tan^3 x\sec^3 x\,dx\).
Here \(k=1\) and \(n=3\). Reserving one \(\sec x\tan x\) leaves \(\tan^2 x\sec^2 x\):
For the check, differentiate and factor the result:
Thus the differentiated expression agrees with the integrand.
General Formula and Its Justification
The method gives a compact formula for every pair of exponents in the stated range. To see why, first transform the integral using \(u=\sec x\), then expand the factor \((u^2-1)^k\) by the binomial theorem. Each resulting term is an ordinary power of \(u\).
Here \(\binom{k}{j}\) is the binomial coefficient. The conditions ensure that the sum is finite and that each denominator \(n+2j\) is positive, so none of the power integrations requires a logarithm.
For the justification, begin with the transformed integral and expand the \(k\)-th power. Since \(k\) is a nonnegative integer, the binomial theorem applies:
In the last step, the power rule for integration applies because \(n-1+2j\ne -1\). Indeed, \(n\geq1\) and \(j\geq0\), so \(n+2j\geq1\). Substituting back \(u=\sec x\) gives the displayed formula.
This substitution argument is legitimate locally wherever \(\sec x\) has a nonzero derivative, and the resulting antiderivative formula can also be checked directly by differentiation wherever \(\cos x\ne0\). The differentiation check confirms the result throughout each interval on which the integrand is defined, including points where \(\tan x=0\).
A Common Substitution Error
A frequent mistake is to reserve \(\sec x\tan x\,dx\) but leave the remaining tangent power unchanged. For example, after writing \(\tan^3 x\sec^2 x\) as \(\tan^2 x\sec x(\sec x\tan x)\), it is incorrect to claim that the remaining factor \(\tan^2 x\sec x\) becomes \(u^2\). It becomes \(u(u^2-1)\), because \(\tan^2 x=u^2-1\).
This wrong replacement would lead to the proposed answer \(\sec^3 x/3+C\). Differentiating exposes the error:
The correction is to convert the remaining tangent power before integrating in \(u\): \(\tan^2 x\sec x=(u^2-1)u\). Integrating \(u(u^2-1)\) gives \(u^4/4-u^2/2\), which differentiates back to the required integrand, as shown in the first worked example.
Why the Method Is Useful
The key decision is structural: an odd tangent power leaves an even number of tangent factors after one \(\tan x\) is paired with a \(\sec x\). Every remaining tangent factor can therefore be handled in pairs through \(\tan^2 x=\sec^2 x-1\). This converts the problem into polynomial algebra rather than requiring a separate trigonometric integration technique for each choice of exponents.
Before applying the method, check both conditions: the tangent exponent must be odd, and there must be at least one secant factor. Then keep the reserved factor visible until the substitution is made. This prevents losing a factor of \(\sec x\) or \(\tan x\). Finally, differentiate the proposed answer. In these integrals, factoring the derivative often reveals \(\sec^2 x-1=\tan^2 x\), making the verification direct.
Check Your Understanding
For each question, identify the substitution structure before carrying out any optional algebra.
- For \(\int \tan^7 x\sec^2 x\,dx\), what are \(k\) and \(n\), and what expression in \(u\) results after \(u=\sec x\)?
- Use the general formula to evaluate \(\int \tan x\sec^3 x\,dx\).
- In \(\int \tan^3 x\sec^4 x\,dx\), which factor should be reserved for \(du\), and what remains to be converted?
- Why does the general formula have no zero denominator when \(k\geq0\) and \(n\geq1\)?
- Differentiate \(\sec^4 x/4-\sec^2 x/2\) and show how \(\sec^2 x-1\) recovers the tangent power.
