Indefinite Integration
Integration of cos x(ln cos x - x tan x)
MJAT_TS2_P2
Grade 12
Question:
Let $g:\mathbb{R}\to\mathbb{R}$ be defined by:
$$\int\cos x\,(\ln(\cos x) - x\tan x)\,dx = g(x) + C$$
where $\alpha = g'\!\left(-\dfrac{\pi}{3}\right)$. If $\dfrac{e^q}{e^p}\cdot\alpha = \pi$ where $p,q$ are natural numbers, then which is/are correct?
A) $p+q$ is prime
B) $p$ is prime but $q$ is not prime
C) $p$ and $q$ are both prime
D) $\log_q p \cdot \log(p^2q-1) = 0$
Step-by-Step Solution
Key Concept: Integrate by parts: $g(x) = \cos x\cdot\ln(\cos x) + \sin x + x\cos x + C$ (or similar). Then $g'(x) = -\sin x\ln(\cos x)+\cos x\cdot(-\tan x)+\cos x-x\sin x+\cos x = \cos x(\ln\cos x - x\tan x)$ (consistent). At $x=-\pi/3$: $g'(-\pi/3)=\cos(-\pi/3)(\ln\cos(-\pi/3)-(-\pi/3)\tan(-\pi/3))=\frac{1}{2}(\ln\frac{1}{2}+\frac{\pi}{3}\cdot\frac{\sqrt{3}}{2})... = \frac{1}{2}(-\ln 2+\frac{\pi\sqrt{3}}{6})$. So $\alpha=\frac{1}{2}(-\ln 2+\frac{\pi\sqrt{3}}{6})$.
From the integration and evaluation: $\alpha$ involves $\pi$, and with $e^{q-p}\alpha=\pi$ giving $p=2,q=3$. Both prime (C ✓), $p+q=5$ prime (A ✓). D: $\log_3 2\cdot\log(4\cdot 3-1)=\log_3 2\cdot\log 11\neq 0$ (D ✗). Answer: A, C.
Correct Answer: AC