Definite Integration
General
Grade 12
Question:
$\int_{0}^{\pi/2} (2\log \sin x - \log \sin 2x) dx$ equals -
$\pi \log 2$
$-\pi \log 2$
$(\pi/2) \log 2$
$-(\pi/2) \log 2$
Step-by-Step Solution
Key Concept: General
$$I = \int_{0}^{\pi/2} (2\log \sin x - \log 2\sin x \cos x) dx = \int_{0}^{\pi/2} (2\log \sin x - \log 2 - \log \sin x - \log \cos x) dx$$ $$= \int_{0}^{\pi/2} \log \sin x dx - \int_{0}^{\pi/2} \log 2 dx - \int_{0}^{\pi/2} \log \cos x dx = -(\pi/2) \log 2$$
Correct Answer: D