AP EAMCET · Maths · Definite Integration
If \(I_1=\int_0^{\pi / 2} \frac{x}{\sin x} d x\), and \(I_2=\int_0^1 \frac{\tan ^{-1} x}{x} d x\), then \(I_1: I_2\) is
- A \(1: 1\)
- B \(2: 1\)
- C \(3: 1\)
- D \(4: 1\)
Answer & Solution
Correct Answer
(B) \(2: 1\)
Step-by-step Solution
Detailed explanation
Given, \(I_1=\int_0^{\pi / 2} \frac{x}{\sin x} d x\) \[ I_2=\int_0^1 \frac{\tan ^{-1} x}{x} d x=\int_0^{\pi / 4} \frac{t}{\tan t} \cdot \sec ^2 t d t \] Put, \(\tan ^{-1} x=t\)…
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