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UPSC NDA 2 2021 Mathematics Solved Paper

Question : 89 of 120

Marks: +1, -0

What is

\int\limits_{0}^{\pi} \ln\left(\tan\frac{x}{2}\right)\,dx

equal to?

Solution:

Concept:

Use the property

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

together with the identity

\tan(\pi - \theta) = -\tan\theta

to simplify the integral.

Explanation:

Let

I = \int_0^{\pi} \ln\left(\tan\frac{x}{2}\right) dx

Apply the property with

a = \pi

I = \int_0^{\pi} \ln\left(\tan\left(\frac{\pi - x}{2}\right)\right) dx

Since

\frac{\pi - x}{2} = \frac{\pi}{2} - \frac{x}{2}

, we have

\tan\left(\frac{\pi}{2} - \frac{x}{2}\right) = \cot\frac{x}{2}

Thus

I = \int_0^{\pi} \ln\left(\cot\frac{x}{2}\right) dx = \int_0^{\pi} \ln\left(\frac{1}{\tan\frac{x}{2}}\right) dx = -\int_0^{\pi} \ln\left(\tan\frac{x}{2}\right) dx = -I

Hence

I = -I

, so

2I = 0

and

I = 0

(Note: The integral is convergent; the integrand has a singularity at

x=0

and

x=\pi

, but the symmetry cancels the area.)

Answer:

0

(Option A)

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