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Tangent and Secant Powers · Tutorial 148 of 998

Rewriting Mixed Trigonometric Factors Before Integration

Learn to rewrite selected squared factors in mixed trigonometric integrals, choose the matching substitution, and verify the result by differentiation.

Intermediate 11 min read

What You'll Learn

  • Use the squared identities to replace tangent or cotangent factors with secant or cosecant expressions.
  • Identify when a rewritten integrand contains the derivative factor for secant, tangent, cosecant, or cotangent.
  • Track the negative signs in substitutions involving cosecant and cotangent.
  • Integrate rewritten products by substitution and verify each antiderivative by differentiation.
  • Recognize when a proposed rewrite does not produce a useful substitution.

Why Rewrite a Trigonometric Product?

A product of tangent and secant factors, or cotangent and cosecant factors, may not initially display the derivative needed for a simple substitution. The factors are related, however, by squared identities. Replacing a selected squared factor can make the rest of the integrand a polynomial in one trigonometric function, while leaving its derivative factor in place.

The key is to make the rewrite serve a specific substitution. For example, the derivative of \(\sec x\) is \(\sec x\tan x\), and the derivative of \(\csc x\) is \(-\csc x\cot x\). When either of these factor pairs occurs, an appropriate squared identity can express the remaining factors using just \(\sec x\) or \(\csc x\).

Core strategy: Look for a derivative factor, preserve it as a whole, and rewrite other squared factors using an identity so that the remaining expression is in the substitution variable.

The Identities and Derivative Factors

The Pythagorean identity \(\sin^2 x+\cos^2 x=1\), divided by \(\cos^2 x\) or by \(\sin^2 x\), gives the identities used here. They apply wherever the relevant trigonometric functions are defined.

$$ \sec^2 x=1+\tan^2 x, \qquad \tan^2 x=\sec^2 x-1, $$
$$ \csc^2 x=1+\cot^2 x, \qquad \cot^2 x=\csc^2 x-1. $$

The derivative factors to recognize are:

$$ \frac{d}{dx}(\sec x)=\sec x\tan x, \qquad \frac{d}{dx}(\tan x)=\sec^2 x, $$
$$ \frac{d}{dx}(\csc x)=-\csc x\cot x, \qquad \frac{d}{dx}(\cot x)=-\csc^2 x. $$

The negative signs in the last two derivatives matter. When using \(u=\csc x\) or \(u=\cot x\), the differential includes a minus sign. A valid rewrite must preserve the original integrand, including its sign, before that differential is substituted.

A Reliable Rewriting Process

1
Find a candidate derivative factor.
For a secant substitution, look for \(\sec x\tan x\,dx\); for tangent, look for \(\sec^2 x\,dx\). For cosecant, look for \(\csc x\cot x\,dx\); for cotangent, look for \(\csc^2 x\,dx\).
2
Keep that factor together.
Set it aside as the differential part of the substitution. Do not change its sign unless the substitution itself introduces that sign.
3
Rewrite the remaining squared factors.
Use \(\tan^2 x=\sec^2 x-1\) when using secant, or \(\cot^2 x=\csc^2 x-1\) when using cosecant. The identities involving \(1+\tan^2 x\) or \(1+\cot^2 x\) can be useful for tangent or cotangent substitutions.
4
Substitute, integrate, and check.
After substitution, integrate the resulting polynomial or other simple expression. Differentiate the final answer to verify it on an interval where the original integrand is defined.

Worked Example: Rewriting for a Secant Substitution

Worked Example: Integrating \(\tan^3 x\sec^4 x\)

Evaluate \(\int \tan^3 x\sec^4 x\,dx\). The factor \(\sec x\tan x\,dx\) suggests \(u=\sec x\). Separate it without changing the integrand:

$$ \tan^3 x\sec^4 x\,dx =\tan^2 x\sec^3 x\,(\sec x\tan x\,dx). $$

Now use \(\tan^2 x=\sec^2 x-1\):

$$ \int \tan^3 x\sec^4 x\,dx =\int (\sec^2 x-1)\sec^3 x\,(\sec x\tan x\,dx). $$

Let \(u=\sec x\), so \(du=\sec x\tan x\,dx\). Then

$$ \int (u^2-1)u^3\,du =\int (u^5-u^3)\,du =\frac{u^6}{6}-\frac{u^4}{4}+C. $$

Returning to \(x\), an antiderivative is \(\frac{\sec^6 x}{6}-\frac{\sec^4 x}{4}+C\).

To check, differentiate:

$$ \frac{d}{dx}\left(\frac{\sec^6 x}{6}-\frac{\sec^4 x}{4}\right) =\sec^5 x(\sec x\tan x)-\sec^3 x(\sec x\tan x) =\sec^4 x\tan x(\sec^2 x-1) =\sec^4 x\tan^3 x. $$

The last equality uses \(\sec^2 x-1=\tan^2 x\), giving the original integrand.

Worked Example: Rewriting for a Cosecant Substitution

Worked Example: Integrating \(\cot^3 x\csc^2 x\)

Evaluate \(\int \cot^3 x\csc^2 x\,dx\). For \(u=\csc x\), the differential is \(du=-\csc x\cot x\,dx\). Factor the integrand so that this pair is present:

$$ \cot^3 x\csc^2 x\,dx =\cot^2 x\csc x\,(\csc x\cot x\,dx). $$

Use \(\cot^2 x=\csc^2 x-1\). Before substituting, the integral is therefore

$$ \int (\csc^2 x-1)\csc x\,(\csc x\cot x\,dx). $$

Since \(\csc x\cot x\,dx=-du\), the correct substitution is

$$ -\int (u^2-1)u\,du =-\int (u^3-u)\,du =-\frac{u^4}{4}+\frac{u^2}{2}+C. $$

Thus an antiderivative is \(-\frac{\csc^4 x}{4}+\frac{\csc^2 x}{2}+C\). To check, differentiate it:

$$ (-\csc^3 x+\csc x)(-\csc x\cot x) =\csc^2 x\cot x(\csc^2 x-1) =\csc^2 x\cot x\cot^2 x =\cot^3 x\csc^2 x. $$

This agrees with the original integrand. The negative sign from \(du=-\csc x\cot x\,dx\) is essential.

Worked Example: Using the Cotangent Substitution

Worked Example: Integrating \(\cot^2 x\csc^4 x\)

Evaluate \(\int \cot^2 x\csc^4 x\,dx\). The factor \(\csc^2 x\,dx\) suggests \(u=\cot x\), because \(du=-\csc^2 x\,dx\). Keep that factor separate and rewrite the other cosecant square using \(\csc^2 x=1+\cot^2 x\):

$$ \int \cot^2 x\csc^4 x\,dx =\int \cot^2 x\csc^2 x\,(\csc^2 x\,dx) =-\int u^2(1+u^2)\,du. $$

Integrate term by term:

$$ -\int (u^2+u^4)\,du =-\frac{u^3}{3}-\frac{u^5}{5}+C. $$

Returning to \(x\), the result is \(-\frac{\cot^3 x}{3}-\frac{\cot^5 x}{5}+C\). For the check, differentiate:

$$ (-\cot^2 x-\cot^4 x)(-\csc^2 x) =\cot^2 x\csc^2 x(1+\cot^2 x) =\cot^2 x\csc^4 x. $$

The final equality follows from \(1+\cot^2 x=\csc^2 x\), so differentiation recovers the integrand.

Why the Rewriting Works

The method is a direct consequence of substitution. If \(u=g(x)\), then \(du=g'(x)\,dx\). When the integrand can be written as \(F(g(x))g'(x)\,dx\), the substitution changes the integral to \(\int F(u)\,du\). The squared identities help produce this form: they express repeated tangent or cotangent factors in terms of the function selected for \(u\).

For instance, in the first example, the identity changes \(\tan^2 x\) into \(\sec^2 x-1\), while the remaining \(\sec x\tan x\,dx\) is exactly \(d(\sec x)\). In the second, \(\cot^2 x\) becomes \(\csc^2 x-1\), and the remaining \(\csc x\cot x\,dx\) is \(-d(\csc x)\). The substitution is useful because the rewritten expression depends only on \(u\), apart from the differential and its sign.

This reasoning also gives a practical test: after choosing a substitution and applying the identities, check that no unwanted \(x\)-dependent factor remains. If one does remain, that particular rewrite has not yet produced a single-variable integral.

Common Pitfall: Losing the Negative Sign

A frequent error is to substitute \(u=\csc x\) but use \(du=\csc x\cot x\,dx\). The correct derivative is \(du=-\csc x\cot x\,dx\). For the second example, omitting the minus sign would give the incorrect candidate \(\frac{\csc^4 x}{4}-\frac{\csc^2 x}{2}\). At \(x=\pi/4\), the original integrand is \(\cot^3(\pi/4)\csc^2(\pi/4)=2\). The derivative of that incorrect candidate at the same point is \(-2\), not \(2\).

Sign check: For \(u=\csc x\), write \(\csc x\cot x\,dx=-du\). For \(u=\cot x\), write \(\csc^2 x\,dx=-du\). Carry the negative sign through the integral before integrating.

Choosing a Useful Identity

The identities are equivalent, but they are not equally convenient for every substitution. If the intended variable is \(\sec x\), rewriting \(\tan^2 x\) as \(\sec^2 x-1\) usually makes the remaining expression a polynomial in \(\sec x\). If the intended variable is \(\csc x\), rewriting \(\cot^2 x\) as \(\csc^2 x-1\) serves the same role. For a tangent substitution, \(\sec^2 x=1+\tan^2 x\) can express a remaining secant square in tangent; for a cotangent substitution, use \(\csc^2 x=1+\cot^2 x\).

Not every mixed product has the right derivative factor or enough squared factors to make one of these rewrites useful. The identity does not guarantee that an integral becomes simple. It is a tool for reorganizing factors, not a reason to force a substitution that leaves the integrand more complicated.

These rewrites matter because they make the structure of a trigonometric product visible. A product that first appears to involve several functions can, after a careful identity and substitution, become an ordinary polynomial integral. A reliable solution records the factorization, the identity, and the differential explicitly; these steps make both the calculation and its sign easy to audit.

Check Your Understanding

For each question, identify the relevant identity or differential factor before carrying out the calculation.

  1. For \(\int \tan^3 x\sec^4 x\,dx\), which squared factor is rewritten to use \(u=\sec x\)?
  2. If \(u=\csc x\), what is \(du\), including its sign?
  3. Rewrite \(\cot^2 x\) in terms of \(\csc x\), then state which factor in \(\int \cot^3 x\csc^2 x\,dx\) is used as the substitution differential.
  4. For \(\int \cot^2 x\csc^4 x\,dx\), explain why \(u=\cot x\) is a useful choice and state the sign in \(du\).
  5. After a trigonometric substitution, what should you check to confirm that the rewrite has produced an integral in \(u\) alone?

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