Odd Secant Powers With A Tangent Factor

Why This Substitution Fits

When an integral contains both tangent and secant, a useful first question is whether the integrand contains the derivative of one of those functions as a factor. Since the derivative of \(\sec x\) is \(\sec x\tan x\), an integral with a tangent factor and a positive power of secant is a natural place to try \(u=\sec x\).

For the particular family considered here, the remaining secant power becomes a simple power of \(u\). The identity \(\tan^2 x=\sec^2 x-1\) also gives a polynomial expression in \(u\). That expression is useful for understanding the algebra, even though the direct substitution is usually shorter.

The Substitution and Its Conditions

Set \(u=\sec x\). Differentiating gives \(du=\sec x\tan x\,dx\). The integrand can be arranged to display exactly this differential:

Therefore, replacing \(\sec x\) by \(u\) and \(\sec x\tan x\,dx\) by \(du\) turns the integral into a power integral. The power rule applies because \(2k\ne -1\); indeed, \(k\ge 0\), so \(2k+1\) is positive.

The substitution is valid locally on any interval where \(\sec x\) is defined. The resulting antiderivative is also valid on each such interval. One should not treat an interval containing a point where \(\cos x=0\) as a single domain for this formula: the original integrand is undefined there.

Why the Substitution Works

The key is not merely that \(u\) equals secant; it is that the integrand contains the full differential of secant. In the expression \(\sec^{2k+1}x\tan x\,dx\), one secant factor pairs with \(\tan x\,dx\) to make \(du\). Exactly \(2k\) secant factors remain, and these become \(u^{2k}\).

This reasoning also proves the general formula: differentiating its right-hand side gives

The derivative agrees with the integrand wherever \(\cos x\ne0\). Thus the formula is verified, not merely suggested by the substitution.

Where the Polynomial Expansion Appears

The identity \(\sec^2x=1+\tan^2x\) gives another way to see the expression that appears after substituting. Write the integrand as \(\sec^{2k}x(\sec x\tan x\,dx)\), then express the remaining even power as \((1+\tan^2x)^k\). With \(u=\sec x\), the identity \(\tan^2x=\sec^2x-1\) changes that factor into \((1+(u^2-1))^k\).

If expanded by the binomial theorem before simplifying, this is the polynomial

The polynomial equals \(u^{2k}\), because the sum is the binomial expansion of \((1+(u^2-1))^k\). In this problem the terms simplify to one power; the expansion is not a different integration method. It makes clear how the tangent-square identity translates the remaining secant power into a polynomial in the substitution variable.

Worked Examples

A Common Mistake: Losing the Secant in the Differential

A frequent error is to use \(du=\tan x\,dx\) after setting \(u=\sec x\). That differential is incorrect: differentiating \(\sec x\) produces an extra factor of \(\sec x\). For example, a wrong calculation for the three-factor case might be

The proposed answer fails the derivative check:

The correction is to use the actual differential, \(du=\sec x\tan x\,dx\), and take one secant factor out of \(\sec^3x\):

This gives \(\sec^3x/3+C\), whose derivative is exactly the original integrand. Whenever a substitution is chosen, write its differential explicitly before replacing any part of the integrand.

Summary and Use

For every nonnegative integer \(k\), the integral of one tangent factor times \(\sec^{2k+1}x\) reduces to a power integral after \(u=\sec x\). The factor \(\sec x\tan x\,dx\) supplies \(du\); the remaining \(\sec^{2k}x\) becomes \(u^{2k}\). The identity \(\tan^2x=\sec^2x-1\) gives the equivalent polynomial \((1+(u^2-1))^k\), which simplifies to the same power.

The practical checks are simple: preserve the full factor \(\sec x\tan x\,dx\) when making the substitution, and differentiate the final expression to confirm that every power of secant matches.