Why Exponent Parity Matters
A product of powers of tangent and secant can often be simplified before integration. The key is to notice which factor can provide a useful differential, or which even power can be rewritten using the Pythagorean identity. Different exponent patterns favor different approaches.
For instance, when the secant exponent is even, a factor of \(\sec^2 x\,dx\) can be paired with the derivative of \(\tan x\). When the tangent exponent is even, its powers can instead be converted into powers of secant. Choosing between these approaches is not merely a matter of algebraic preference: a good choice can turn the integral into a polynomial integral or a short combination of familiar secant integrals.
The Two Main Reductions
The identity \(\sec^2 x=1+\tan^2 x\), equivalently \(\tan^2 x=\sec^2 x-1\), and the derivative \(\frac{d}{dx}\tan x=\sec^2 x\) give two principal strategies.
This method requires an even secant exponent of at least two, so that a factor \(\sec^2 x\) is available. The exponent \(m\) may be even or odd. The reduction works because the differential \(du\) absorbs exactly the reserved \(\sec^2 x\); the remaining secant powers become powers of \(1+u^2\).
The two reductions can overlap. If both exponents are even, either formula may be available; compare the amount of expansion each requires. If the tangent exponent is odd and the secant exponent is odd, neither of these two rules applies directly in its stated form. A useful additional step is to remove one factor of \(\tan x\sec x\), since \(\frac{d}{dx}\sec x=\sec x\tan x\).
Why the Substitution Works
Here is the key reasoning behind the even-secant reduction. Suppose \(n=2q\) and set \(u=\tan x\). The derivative of the substitution is \(du=\sec^2 x\,dx\), so separate that factor from the original product. The remaining secant power is \(\sec^{2q-2}x=(\sec^2 x)^{q-1}\). Since \(\sec^2 x=1+\tan^2 x\), this remaining factor becomes \((1+u^2)^{q-1}\), while \(\tan^m x\) becomes \(u^m\). Thus the original integrand, including \(dx\), becomes exactly \(u^m(1+u^2)^{q-1}\,du\). No factor is discarded or added.
Because \(q-1\) is a nonnegative integer, the binomial theorem gives a finite expansion. In particular, \[ u^m(1+u^2)^{q-1} =\sum_{k=0}^{q-1}\binom{q-1}{k}u^{m+2k}. \] Therefore, \[ I_{m,2q} =\sum_{k=0}^{q-1}\binom{q-1}{k} \frac{u^{m+2k+1}}{m+2k+1}+C, \qquad u=\tan x. \] The denominators are nonzero because \(m\ge0\) and \(k\ge0\).
Worked Examples
Worked Example: Even Secant Exponent
Evaluate \(\int \tan^3 x\,\sec^4 x\,dx\).
The secant exponent is \(4=2q\), so \(q=2\). Separate one factor of \(\sec^2 x\): \[ \int \tan^3 x\,\sec^4 x\,dx =\int \tan^3 x\,\sec^2 x\,\sec^2 x\,dx. \] Set \(u=\tan x\). Then \(du=\sec^2 x\,dx\), and the remaining factor satisfies \(\sec^2 x=1+u^2\). Hence \[ \int \tan^3 x\,\sec^4 x\,dx =\int u^3(1+u^2)\,du =\int (u^3+u^5)\,du. \] Integrating term by term gives \[ \frac{u^4}{4}+\frac{u^6}{6}+C =\frac{\tan^4 x}{4}+\frac{\tan^6 x}{6}+C. \]
Check by differentiating: \[ \frac{d}{dx}\left(\frac{\tan^4 x}{4}+\frac{\tan^6 x}{6}\right) =\tan^3 x\,\sec^2 x+\tan^5 x\,\sec^2 x =\tan^3 x\,\sec^2 x(1+\tan^2 x) =\tan^3 x\,\sec^4 x. \] The derivative is the original integrand.
Worked Example: Even Tangent Exponent and Secant Recurrence
Evaluate \(\int \tan^4 x\,\sec^3 x\,dx\).
Since \(\tan^4 x=(\sec^2 x-1)^2\), expand first: \[ \int \tan^4 x\,\sec^3 x\,dx =\int(\sec^4 x-2\sec^2 x+1)\sec^3 x\,dx =\int\sec^7 x\,dx-2\int\sec^5 x\,dx+\int\sec^3 x\,dx. \]
For odd \(r>1\), the secant-power reduction is \[ \int\sec^r x\,dx =\frac{\sec^{r-2}x\tan x}{r-1} +\frac{r-2}{r-1}\int\sec^{r-2}x\,dx. \] Using it successively, and writing \(L=\ln|\sec x+\tan x|\), gives \[ \int\sec^3 x\,dx=\frac12\sec x\tan x+\frac12L, \] \[ \int\sec^5 x\,dx=\frac14\sec^3 x\tan x+\frac38\sec x\tan x+\frac38L, \] and \[ \int\sec^7 x\,dx=\frac16\sec^5 x\tan x+\frac5{24}\sec^3 x\tan x+\frac5{16}\sec x\tan x+\frac5{16}L. \] Combining these with the coefficients \(1,-2,1\) yields \[ \int \tan^4 x\,\sec^3 x\,dx =\frac16\sec^5 x\tan x-\frac7{24}\sec^3 x\tan x +\frac1{16}\sec x\tan x+\frac1{16}L+C. \]
To check, differentiate the expression. The needed derivatives are \[ \frac{d}{dx}(\sec^5x\tan x)=6\sec^7x-5\sec^5x,\quad \frac{d}{dx}(\sec^3x\tan x)=4\sec^5x-3\sec^3x, \] \[ \frac{d}{dx}(\sec x\tan x)=2\sec^3x-\sec x,\quad \frac{dL}{dx}=\sec x. \] Thus the derivative of the proposed answer is \[ \sec^7x-2\sec^5x+\sec^3x =(\sec^2x-1)^2\sec^3x =\tan^4x\,\sec^3x, \] as required.
Worked Example: Both Exponents Odd
Evaluate \(\int \tan^3 x\,\sec^3 x\,dx\).
Use \(\tan^2 x=\sec^2 x-1\) to rewrite the integrand while retaining one factor of \(\tan x\): \[ \tan^3x\,\sec^3x =\tan x(\tan^2x)\sec^3x =\tan x(\sec^2x-1)\sec^3x =\tan x(\sec^5x-\sec^3x). \] Now set \(u=\sec x\), so \(du=\sec x\tan x\,dx\). Factor the integrand to expose this differential: \[ \int \tan x(\sec^5x-\sec^3x)\,dx =\int \sec x\tan x(\sec^4x-\sec^2x)\,dx =\int(u^4-u^2)\,du. \] Therefore, \[ \int \tan^3x\,\sec^3x\,dx =\frac{u^5}{5}-\frac{u^3}{3}+C =\frac{\sec^5x}{5}-\frac{\sec^3x}{3}+C. \]
Differentiate to check: \[ \frac{d}{dx}\left(\frac{\sec^5x}{5}-\frac{\sec^3x}{3}\right) =\sec^5x\tan x-\sec^3x\tan x =\sec^3x\tan x(\sec^2x-1) =\sec^3x\tan^3x. \] This is the original integrand.
Choosing a Route and Avoiding a Common Error
The exponent pattern suggests a route, but it does not remove the need to track every factor. The following summary distinguishes the main options.
| Exponent pattern | Useful step | Typical result |
|---|---|---|
| Secant exponent even and at least two | Reserve \(\sec^2 x\,dx\); set \(u=\tan x\) | A polynomial in \(u\), after expansion |
| Tangent exponent even | Replace \(\tan^2x\) by \(\sec^2x-1\) | A sum of secant-power integrals |
| Both exponents odd | Retain \(\tan x\sec x\,dx\); set \(u=\sec x\) | A polynomial in \(u\), after rewriting the remaining tangent powers |
A common mistake is to use \(u=\tan x\) but forget that the substitution absorbs only one factor of \(\sec^2 x\,dx\). For example, an incorrect step for \(\int \tan^3x\sec^4x\,dx\) is \[ \int \tan^3x\sec^4x\,dx \overset{\text{wrong}}{=} \int u^3(1+u^2)^2\,du. \] That expression converts all four secant factors into powers of \(1+u^2\) but does not reserve the factor that supplies \(du\). The correct separation is \(\sec^4x=\sec^2x\sec^2x\): one \(\sec^2x\,dx\) becomes \(du\), and the remaining \(\sec^2x\) becomes \(1+u^2\). The correct transformed integrand is \(u^3(1+u^2)\), not \(u^3(1+u^2)^2\).
Why These Reductions Are Useful
These methods turn a mixed trigonometric product into a form with a clear endpoint: a polynomial integral after substitution, or a finite combination of secant powers. This is especially useful when direct integration by parts would create more complicated products instead of simplifying the exponents.
It is also important to apply the conditions precisely. The substitution \(u=\tan x\) in the displayed reduction requires a factor \(\sec^2 x\,dx\), so it applies there when \(n\) is even and \(n\ge2\). Rewriting an even tangent power does not require an even secant exponent. When both exponents are odd, retaining \(\tan x\sec x\,dx\) can expose the derivative of secant. In each case, the goal is to transform the integrand without changing it, then verify the result by differentiation.
Check Your Understanding
For each question, identify the useful exponent pattern before calculating.
- For \(\int\tan^2x\,\sec^4x\,dx\), which of the two main reductions is available? What substitution follows from reserving \(\sec^2x\,dx\)?
- Rewrite \(\tan^6x\,\sec^3x\) as a finite sum of secant powers, without evaluating the resulting integrals.
- Explain why the substitution \(u=\tan x\) is useful for an even secant exponent of at least two.
- For \(\int\tan^3x\,\sec^5x\,dx\), identify a factor that can provide \(du\) for \(u=\sec x\).
- Differentiate \(\frac14\tan^4x+\frac16\tan^6x\) and compare the result with the integrand in the first worked example.
