WBJEE · Maths · Definite Integration
Let \(I=\int_{0}^{{100 \pi }} \sqrt{(1-\cos 2 x)} d x\), then
- A \(I=0\)
- B \(I=200 \sqrt{2}\)
- C \(I=\pi \sqrt{2}\)
- D \(I=100\)
Answer & Solution
Correct Answer
(B) \(I=200 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(I=\int_{0}^{1004} \sqrt{1-\cos 2 x} d x\) \(=\int_{0}^{100 \pi} \sqrt{2 \sin ^{2} x} d x\) \(=\sqrt{2} \int_{0}^{100 \pi} \sin x \mid d x\) \(=\sqrt{2} \times 100 \int_{0}^{n} \sin x|d x|\) \([\sin x \mid\) has period of \(\pi]\) \(=100 \sqrt{2} \int_{0}^{\pi} sin x d x\)…
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