TS EAMCET · Maths · Definite Integration
\(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta\) is equal to
- A \(0\)
- B \(1\)
- C \(2\)
- D \(-1\)
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
Let \(I=\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta\) Let \(\quad f(\theta)=\log \left(\frac{2-\sin \theta}{2+\sin \theta}\right)\)…
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