COMEDK · Maths · 30. Definite Integration
\(\int_0^\pi \dfrac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x\) is equal to
- A \(\dfrac{\pi}{2}\)
- B \(\pi\)
- C \(2 \pi\)
- D \(\dfrac{\pi}{4}\)
Answer & Solution
Correct Answer
(A) \(\dfrac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
Let \(I = \int_0^{\pi} \dfrac{e^{\cos x}}{e^{\cos x} + e^{-\cos x}} dx\). Using the property \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\), we have: \(I = \int_0^{\pi} \dfrac{e^{\cos(\pi-x)}}{e^{\cos(\pi-x)} + e^{-\cos(\pi-x)}} dx\) Since \(\cos(\pi-x) = -\cos x\), the integral…
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