COMEDK · Maths · 30. Definite Integration
\(\int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \dfrac{1}{1+\sqrt{\tan x}} d x=\)
- A \(\dfrac{\pi}{3}\)
- B \(\dfrac{\pi}{2}\)
- C \(\dfrac{\pi}{6}\)
- D \(\dfrac{\pi}{12}\)
Answer & Solution
Correct Answer
(D) \(\dfrac{\pi}{12}\)
Step-by-step Solution
Detailed explanation
Let \(I = \int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \dfrac{1}{1 + \sqrt{\tan x}} dx\). Using the property \(\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx\), we have \(a + b = \dfrac{\pi}{6} + \dfrac{\pi}{3} = \dfrac{\pi}{2}\).…
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