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\( \int_{\frac{\pi}{2}}^{\pi / 2} \frac{1}{1+e^{\sin x}} d x \) |
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Answer» -π/2∫π/2 \(\frac1{1+e^{sin\,x}}dx\) = I (Let) ...(1) ⇒ I = -π/2∫π/2 \(\frac1{1+e^{sin\,x}}dx\) = -π/2∫π/2 \(\frac{e^{sin\,x}}{1+e^{sin\,x}}dx\) ...(2) By adding equations (1) and (2), we get 2I = -π/2∫π/2 \(\frac{1+e^{sin\,x}}{1+e^{sin\,x}}dx\) \(=[x]^{\pi/2}_{-\pi/2}\) = π/2 + π/2 = π ⇒ I = π/2 Hence, -π/2∫π/2 \(\frac1{1+e^{sin\,x}}dx\) = π/2. |
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