YouFind the following indefinite integral:integral e to the power of 3 x end exponent cos left parenthesis e to the power of 3 x end exponent right parenthesis d xA. 3 e to the power of 3 x end exponent sin left parenthesis e to the power of 3 x end exponent right parenthesis plus cB. 1 third sin left parenthesis e to the power of 3 x end exponent right parenthesis plus cC. 3 sin left parenthesis e to the power of 3 x end exponent right parenthesis plus cD. 1 third e to the power of 3 x end exponent sin left parenthesis e to the power of 3 x end exponent right parenthesis plus c

The integral you're asking about is a classic example of an integral that can be solved using the method of integration by substitution. Here are the steps to solve it:

1. Identify a part of the integral that you could substitute for a single variable to simplify the integral. In this case, we can let u = e^(3x). Then, du = 3e^(3x) dx.

2. Rewrite the integral in terms of u. The integral becomes ∫ cos(u) du/3.

3. Now, the integral is much simpler to solve. The integral of cos(u) is sin(u), so ∫ cos(u) du/3 = 1/3 sin(u) + C.

4. Substitute e^(3x) back in for u to get the final answer: 1/3 sin(e^(3x)) + C.

So, the correct answer is B. 1 third sin left parenthesis e to the power of 3 x end exponent right parenthesis plus c.