If $\int_a^b \sin x dx = 8$ and $\int_0^{a+b} \cos x dx = 9$, then:
Step-by-Step Solution
Key Concept: Recognize that integrals of periodic trigonometric functions over specified intervals yield linear equations in $a$ and $b$ that can be solved simultaneously.
Using the given property that $\int_a^b |\sin x| dx = 8 - (b-a) = \frac{8\pi}{2}$ when the period is $\pi$, we have $b - a = \frac{8\pi}{2}$ from equation (1). Similarly, $\int_0^a \cos x dx = 9$ gives $a + b - 0 = \frac{9\pi}{2}$ from equation (2). Solving: $a = \frac{\pi}{4}$ and $b = \frac{17\pi}{4}$. Then $|a+b| = \frac{9\pi}{2}$, $|a-b| = 4\pi = \frac{17}{b}$, and $\int_a^b \sec^2 x dx = [\tan x]_a^b = 0$.
Correct Answer: 2,3