Integrating tan 3x · sec 3x: A Step‑by‑Step Guide
Once 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 And it works..
Short version: it depends. Long version — keep reading.
Introduction
The function tan 3x sec 3x combines two trigonometric functions with the same argument, 3x. The key insight is that the derivative of sec 3x is 3 sec 3x tan 3x. By recognizing this relationship, the integral collapses to a simple logarithm. 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 Turns out it matters..
Why This Works: A Deeper Look
The integral hinges on the identity
[ \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. In our case, (u = 3x) gives (\frac{du}{dx} = 3). The presence of this constant factor is why we had to divide by 3 in the substitution step. 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 The details matter here..
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? | If the argument is a nonlinear function, say (\tan(g(x))\sec(g(x))), you can still substitute (u = \sec(g(x))). The differential will involve (g'(x)), so you may need to factor it out or use integration by parts. That said, |
| **Can we integrate (\tan x \sec x) directly? Also, ** | Yes. With (u = \sec x), the integral becomes (\int \tan x \sec x , dx = \sec x + C). Still, |
| **Why do we divide by 3? ** | Because (du = 3 \sec 3x \tan 3x,dx). Consider this: to isolate (dx) in terms of (du), we divide by 3. |
| What if the integrand were (\tan 3x \sec^2 3x)? | Use substitution (u = \tan 3x). So then (du = 3 \sec^2 3x,dx), leading to (\int \tan 3x \sec^2 3x,dx = \frac{1}{3}\tan^2 3x + C). |
| **Is this approach always the best?Which means ** | For products of (\tan) and (\sec) with the same argument, yes. In real terms, other methods (e. Because of that, g. , 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. Now, by setting (u = \sec 3x) and noting that (du = 3 \sec 3x \tan 3x,dx), the integral collapses to a constant times (u). Now, this technique not only provides a quick solution but also deepens your understanding of the interplay between trigonometric functions and their derivatives. 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 Simple as that..
