Question

Use a computer algebra system to evaluate the following indefinite integrals. Assume that a is a positive real number. $$\int \tan ^{2} 3 x d x$$

Short answer

#Question# Evaluate the indefinite integral of the function: $$\int \tan ^{2} 3 x d x$$ #Answer# $$\int \tan ^{2} 3 x d x = \frac{1}{3}\tan(3x) - x + C$$

Step-by-step solution

Rewrite the integrand using trigonometric identities

In order to integrate the given function, we can rewrite \(\tan^2(3x)\) using the following trigonometric identity: $$\tan^2(\theta) = \sec^2(\theta) - 1$$ Applying this identity with \(\theta=3x\), we get: $$\tan^2(3x) = \sec^2(3x) - 1$$ Now, the integral becomes: $$\int \tan^2(3x) dx = \int (\sec^2(3x) - 1) dx$$

Integrate the two terms separately

Now we can split the integral into two separate integrals as follows: $$\int (\sec^2(3x) - 1) dx = \int \sec^2(3x) dx - \int 1 dx$$ Integrating \(\sec^2(3x)\) with respect to \(x\), we obtain $$\frac{1}{3}\tan(3x) + C_1$$. Integrating \(1\) with respect to \(x\), we obtain \(x + C_2\). Hence the integral can be written as: $$\int (\sec^2(3x) - 1) dx = \frac{1}{3}\tan(3x) + C_1 - x - C_2$$ Combine \(C_1\) and \(C_2\) into a single constant \(C\), and we get the final answer: $$\int (\sec^2(3x) - 1) dx = \frac{1}{3}\tan(3x) - x + C$$