MHT CET · Maths · Definite Integration
\(\int_{0}^{\pi} \frac{e^{\cos x}}{\left(e^{\cos x}+e^{-\cos x)}\right.} d x=\)
- A \(\frac{-\pi}{2}\)
- B \(-\pi\)
- C \(\pi\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
\( \text { Let}I =\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x} d x}...(1) \)
\( =\int_{0}^{\pi} \frac{e^{\cos (\pi-x)}}{e^{\cos (\pi-x)}+e^{-\cos (\pi-x)}} d x \)
\( =\int_{0}^{\pi} \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x} d x}...(2)\)
Adding equation (1) \& (2), we get
\(
2 I=\int_{0}^{\pi} 1 d x=[x]_{0}^{\pi} \Rightarrow I=\frac{\pi}{2}
\)
\( =\int_{0}^{\pi} \frac{e^{\cos (\pi-x)}}{e^{\cos (\pi-x)}+e^{-\cos (\pi-x)}} d x \)
\( =\int_{0}^{\pi} \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x} d x}...(2)\)
Adding equation (1) \& (2), we get
\(
2 I=\int_{0}^{\pi} 1 d x=[x]_{0}^{\pi} \Rightarrow I=\frac{\pi}{2}
\)
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