Integral Of Tan 3x Sec 3x

Integrating tan 3x · sec 3x: A Step‑by‑Step Guide

When you first see the integral
[ \int \tan 3x , \sec 3x , dx, ]
it may look intimidating, but it is actually a textbook example of how a clever substitution can turn a seemingly complex expression into something trivial. In this article we’ll walk through the integration process, explain why the trick works, and explore related integrals that share the same underlying strategy.

Introduction

The function tan 3x sec 3x combines two trigonometric functions with the same argument, 3x. In practice, by recognizing this relationship, the integral collapses to a simple logarithm. The key insight is that the derivative of sec 3x is 3 sec 3x tan 3x. This technique—using the derivative of a function that appears inside the integrand—is a cornerstone of integration by substitution and is widely applicable to many trigonometric integrals.

Step 1: Identify the Pattern

The derivative of (\sec u) with respect to (u) is (\sec u \tan u).
If we set (u = 3x), then

[ \frac{d}{dx}\bigl(\sec 3x\bigr) = 3 \sec 3x \tan 3x. ]

Notice that the integrand is exactly (\tan 3x \sec 3x), missing only a factor of 3. This observation suggests a substitution that will eliminate the extra constant.

Step 2: Substitution

Let

[ u = \sec 3x. ]

Then differentiate both sides:

[ du = 3 \sec 3x \tan 3x , dx \quad\Longrightarrow\quad dx = \frac{du}{3 \sec 3x \tan 3x}. ]

On the flip side, we can avoid solving for (dx) by simply recognizing that

[ 3 \sec 3x \tan 3x , dx = du. ]

Thus,

[ \sec 3x \tan 3x , dx = \frac{du}{3}. ]

Substituting into the integral gives

[ \int \tan 3x , \sec 3x , dx = \int \frac{1}{3}, du = \frac{1}{3} \int du. ]

Step 3: Integrate

Integrating (du) is straightforward:

[ \frac{1}{3} \int du = \frac{1}{3} u + C. ]

Finally, revert to the original variable:

[ \frac{1}{3} u + C = \frac{1}{3} \sec 3x + C. ]

So the antiderivative is

[ \boxed{\displaystyle \int \tan 3x , \sec 3x , dx = \frac{1}{3}\sec 3x + C.} ]

Quick Check: Differentiation

To confirm, differentiate the result:

[ \frac{d}{dx}\left(\frac{1}{3}\sec 3x\right) = \frac{1}{3} \cdot 3 \sec 3x \tan 3x = \sec 3x \tan 3x, ]

which matches the original integrand. The solution is correct.

Why This Works: A Deeper Look

The integral hinges on the identity

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[ \frac{d}{dx}\sec u = \sec u \tan u \cdot \frac{du}{dx}. ]

When the argument (u) is a linear function of (x), (\frac{du}{dx}) is a constant. Practically speaking, the presence of this constant factor is why we had to divide by 3 in the substitution step. Practically speaking, in our case, (u = 3x) gives (\frac{du}{dx} = 3). This pattern appears frequently in integrals involving products of (\sec) and (\tan) with the same argument.

Related Integrals

The same substitution technique solves several other integrals that involve (\sec) and (\tan) together. Here are a few examples with brief solutions:

Integral	Substitution	Result
(\displaystyle \int \sec 5x \tan 5x , dx)	(u = \sec 5x)	(\frac{1}{5}\sec 5x + C)
(\displaystyle \int \sin 2x \cos 2x , dx)	(u = \sin 2x)	(\frac{1}{4}\sin^2 2x + C)
(\displaystyle \int \tan^2 x \sec^2 x , dx)	(u = \tan x)	(\frac{1}{3}\tan^3 x + C)
(\displaystyle \int \sec^2 4x , dx)	(u = \tan 4x)	(\frac{1}{4}\tan 4x + C)

In each case, the integrand contains a function whose derivative is also present, multiplied by the derivative of its inner argument.

A Practical Example: Solving an Integral in Context

Suppose you’re studying the motion of a particle whose velocity is given by

[ v(t) = 2 \tan(3t) \sec(3t). ]

To find the displacement over a time interval ([0, T]), you need to integrate the velocity:

[ s(T) = \int_0^T 2 \tan(3t) \sec(3t) , dt. ]

Using the result above,

[ s(T) = 2 \left[\frac{1}{3}\sec(3t)\right]_0^T = \frac{2}{3}\bigl(\sec(3T) - \sec(0)\bigr). ]

Since (\sec(0) = 1), the displacement simplifies to

[ s(T) = \frac{2}{3}\bigl(\sec(3T) - 1\bigr). ]

This simple closed‑form expression would be impossible to guess without recognizing the derivative pattern.

Frequently Asked Questions (FAQ)

Question	Answer
**What if the argument isn’t linear?Worth adding: then (du = 3 \sec^2 3x,dx), leading to (\int \tan 3x \sec^2 3x,dx = \frac{1}{3}\tan^2 3x + C).
**Can we integrate (\tan x \sec x) directly?
What if the integrand were (\tan 3x \sec^2 3x)?Practically speaking,	Because (du = 3 \sec 3x \tan 3x,dx). With (u = \sec x), the integral becomes (\int \tan x \sec x , dx = \sec x + C).
Why do we divide by 3?Because of that,	For products of (\tan) and (\sec) with the same argument, yes. g.**
Is this approach always the best? To isolate (dx) in terms of (du), we divide by 3. The differential will involve (g'(x)), so you may need to factor it out or use integration by parts.	Yes. Other methods (e., trigonometric identities) can also work but typically add unnecessary complexity.

Conclusion

The integral of (\tan 3x \sec 3x) is a textbook illustration of how recognizing the derivative of a function inside an integrand can simplify the problem dramatically. Because of that, this technique not only provides a quick solution but also deepens your understanding of the interplay between trigonometric functions and their derivatives. By setting (u = \sec 3x) and noting that (du = 3 \sec 3x \tan 3x,dx), the integral collapses to a constant times (u). Armed with this insight, you can tackle a wide range of integrals that involve (\tan) and (\sec) together, turning potential stumbling blocks into straightforward calculations.