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COMEDK 2025 Morning Solved Paper

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Question : 159 of 180

Marks: +1, -0

\int_0^\pi \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x

is equal to

$\pi$
$2 \pi$
$\frac{\pi}{4}$
$\frac{\pi}{2}$

Solution:

Concept:

Use the property of definite integrals:

\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx

, and add the two forms to simplify.

Explanation:

Let

I = \int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x} + e^{-\cos x}} \, dx

.

Under the substitution

x \to \pi - x

, we have

\cos(\pi - x) = -\cos x

.

Thus, the integrand becomes

\frac{e^{-\cos x}}{e^{-\cos x} + e^{\cos x}}

.

So

I = \int_{0}^{\pi} \frac{e^{-\cos x}}{e^{\cos x} + e^{-\cos x}} \, dx

as well.

Add the two expressions for

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

.

The numerator simplifies:

e^{\cos x} + e^{-\cos x}

, so the integrand becomes

1

.

Therefore,

2I = \int_{0}^{\pi} 1 \, dx = \pi

.

Hence,

I = \frac{\pi}{2}

.

Answer:

\displaystyle \frac{\pi}{2}

(Option D)

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