Final answer:
To evaluate the given integral f(6√(1-sin(x)))dx, simplify the expression inside the square root, then integrate 6|cos(x)| over one period.
Step-by-step explanation:
To evaluate the integral f(6√(1-sin(x)))dx, we can start by simplifying the expression inside the square root. Using the identity sin^2(x) + cos^2(x) = 1, we can rewrite 1 - sin(x) as cos^2(x). Substituting this back into the integral, we have f(6√(cos^2(x)))dx.
Next, we can simplify the square root by taking out the constant 6. This gives us 6√(cos^2(x)) = 6|cos(x)|.
Finally, we can integrate f(6|cos(x)|)dx. Since the absolute value of cosine is an even function, we can integrate over one period and double the result. The integral of 6|cos(x)|dx over one period is 12.