Q111 / 180 · COMEDK 2025 (10 May Shift 1) · Mathematics
$\int_0^\pi \dfrac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x$ is equal to
- Correct answerA$\dfrac{\pi}{2}$
- B$\pi$
- C$2 \pi$
- D$\dfrac{\pi}{4}$
Solution
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 becomes:
$I = \int_0^{\pi} \dfrac{e^{-\cos x}}{e^{-\cos x} + e^{\cos x}} dx$
Adding the two expressions for $I$:
$2I = \int_0^{\pi} \left( \dfrac{e^{\cos x}}{e^{\cos x} + e^{-\cos x}} + \dfrac{e^{-\cos x}}{e^{\cos x} + e^{-\cos x}} \right) dx$
$2I = \int_0^{\pi} \dfrac{e^{\cos x} + e^{-\cos x}}{e^{\cos x} + e^{-\cos x}} dx$
$2I = \int_0^{\pi} 1 dx = \pi$
$I = \dfrac{\pi}{2}$
Answer: $\dfrac{\pi}{2}$
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 becomes:
$I = \int_0^{\pi} \dfrac{e^{-\cos x}}{e^{-\cos x} + e^{\cos x}} dx$
Adding the two expressions for $I$:
$2I = \int_0^{\pi} \left( \dfrac{e^{\cos x}}{e^{\cos x} + e^{-\cos x}} + \dfrac{e^{-\cos x}}{e^{\cos x} + e^{-\cos x}} \right) dx$
$2I = \int_0^{\pi} \dfrac{e^{\cos x} + e^{-\cos x}}{e^{\cos x} + e^{-\cos x}} dx$
$2I = \int_0^{\pi} 1 dx = \pi$
$I = \dfrac{\pi}{2}$
Answer: $\dfrac{\pi}{2}$
Hence, the correct answer is option: 'A'.