AP EAMCET · Maths · Definite Integration
\(\int_0^{\pi / 2} \frac{\sin x}{1+\cos x+\sin x} d x=\)
- A \(\frac{\pi}{2}+\frac{1}{2} \log 2\)
- B \(\frac{\pi}{4}-\frac{1}{2} \log 2\)
- C \(\frac{\pi}{4}\)
- D \(\frac{3 \pi}{4}+\log 2\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{4}-\frac{1}{2} \log 2\)
Step-by-step Solution
Detailed explanation
\(I=\int_0^{\pi/2} \frac{\sin x}{1+\cos x+\sin x} dx\) \(I=\int_0^{\pi/2} \frac{\sin(\pi/2-x)}{1+\cos(\pi/2-x)+\sin(\pi/2-x)} dx = \int_0^{\pi/2} \frac{\cos x}{1+\sin x+\cos x} dx\)…
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