Evaluate the integral: I = integral dx/(cos²(3x+2))
Evaluate the integral: $$ I = \int \frac{dx}{\cos^2(3x+2)}. $$
Explanation of Method
The reciprocal of $\cos^2$ is $\sec^2$, whose antiderivative is $\tan$. Rewriting the integrand as $\sec^2(3x+2)$ and substituting $u=3x+2$ reduces the problem to the standard $\int \sec^2 u\,du = \tan u$.
- Rewrite $\frac{1}{\cos^2}$ as $\sec^2$.
- Substitute $u$ for the linear argument.
- Integrate using the standard $\sec^2$ antiderivative.
- Substitute back and add the constant of integration.
Step 1 · Recognize the reciprocal identity
To solve the integral, we recognize that:
$$ \frac{1}{\cos^2 u} = \sec^2 u. $$
Thus, we rewrite the integral as:
$$ I = \int \sec^2(3x+2)\,dx. $$
Step 2 · Substituting a new variable
Let:
$$ u = 3x+2. $$
Then, differentiate both sides:
$$ \frac{du}{dx} = 3 \quad \Rightarrow \quad du = 3\,dx \quad \Rightarrow \quad dx = \frac{du}{3}. $$
Step 3 · Transforming the integral
Rewriting the integral in terms of $u$:
$$ I = \int \sec^2 u \cdot \frac{du}{3}. $$
Step 4 · Evaluating the integral
Since $\int \sec^2 u\,du = \tan u$, we obtain:
$$ I = \frac{1}{3}\int \sec^2 u\,du = \frac{1}{3}\tan u. $$
Step 5 · Substituting back $u$
Replacing $u = 3x+2$:
$$ I = \frac{1}{3}\tan(3x+2) + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Rewritten integrand | $\sec^2(3x+2)$ |
| Substitution | $u = 3x+2$ |
| Antiderivative | $\tfrac13\tan u$ |
Additional Notes
- The domain of $I$ excludes points where $\cos(3x+2)=0$, i.e. $3x+2=\tfrac\pi2+k\pi$ for integer $k$.
- Recognizing $\frac{1}{\cos^2 u}=\sec^2 u$ as the derivative of $\tan u$ is the key first step — without it, the integrand doesn't obviously match a standard antiderivative.
- The result can be checked by differentiating $\tfrac13\tan(3x+2)$ with the chain rule and confirming it returns $\dfrac{1}{\cos^2(3x+2)}$.
Result
The solution to the integral is: