Question

\( \displaystyle \int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}\,dx \) is equal to

• A. \( \pi \)
• B. \( 2\pi \)
• C. \( \frac{\pi}{4} \)
• D. \( \frac{\pi}{2} \)

Hint:

For definite integrals from \(0\) to \(\pi\), use the transformation \(x \to \pi-x\), especially when \(\cos x\) appears in the integrand.

Answer 1

The Correct Option is D

Step 1: Let the integral be \(I\).
\[ I=\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}\,dx. \]

Step 2: Use the property of definite integral.
Using the property:
\[ \int_0^\pi f(x)\,dx=\int_0^\pi f(\pi-x)\,dx. \]

Step 3: Replace \(x\) by \(\pi-x\).
\[ I=\int_{0}^{\pi} \frac{e^{\cos(\pi-x)}}{e^{\cos(\pi-x)}+e^{-\cos(\pi-x)}}\,dx. \]
Since:
\[ \cos(\pi-x)=-\cos x, \]
we get:
\[ I=\int_{0}^{\pi} \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}}\,dx. \]

Step 4: Add both forms of \(I\).
\[ 2I=\int_{0}^{\pi} \left[ \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} + \frac{e^{-\cos x}}{e^{\cos x}+e^{-\cos x}} \right]dx. \]

Step 5: Simplify the integrand.
\[ 2I=\int_{0}^{\pi} \frac{e^{\cos x}+e^{-\cos x}}{e^{\cos x}+e^{-\cos x}}\,dx. \]
\[ 2I=\int_{0}^{\pi} 1\,dx. \]

Step 6: Evaluate the integral.
\[ 2I=\left[x\right]_{0}^{\pi}. \]
\[ 2I=\pi. \]

Step 7: Find \(I\).
\[ I=\frac{\pi}{2}. \]
Final Answer:
\[ \boxed{\frac{\pi}{2}} \]