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MHTCET 17 Apr 2026 Shift 1 Paper

Question : 133 of 150

Marks: +1, -0

The value of

\int_{0}^{\pi} \frac{1}{1+2^{\cos x}} d x

\_\_\_\_

Solution:

Concept:

Use the property of definite integrals:

\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx

Then add both forms of the integral to simplify the integrand.

Explanation:

Let

I = \int_{0}^{\pi} \frac{1}{1+2^{\cos x}} \, dx

Apply the property by replacing

x

with

\pi - x

I = \int_{0}^{\pi} \frac{1}{1+2^{\cos(\pi - x)}} \, dx

Since

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

, we get

I = \int_{0}^{\pi} \frac{1}{1+2^{-\cos x}} \, dx

Now,

2^{-\cos x} = \frac{1}{2^{\cos x}}

, so

\frac{1}{1+2^{-\cos x}} = \frac{1}{1+\frac{1}{2^{\cos x}}} = \frac{2^{\cos x}}{1+2^{\cos x}}

Thus,

I = \int_{0}^{\pi} \frac{2^{\cos x}}{1+2^{\cos x}} \, dx

Add the two expressions for

I

2I = \int_{0}^{\pi} \left( \frac{1}{1+2^{\cos x}} + \frac{2^{\cos x}}{1+2^{\cos x}} \right) dx

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

Therefore,

I = \frac{\pi}{2}

Answer:

\frac{\pi}{2}

Option A.

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