JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Definite Integrals

The value of the integral

\int_{1/2}^{2} \frac{t a n ^{- 1} x}{x} d x

is equal to

Answer: (D)

Solution

Let

I = \int_{1/2}^{2} \frac{tan ^{- 1} x}{x} d x

. The substitution

x = \frac{1}{t}

maps the limits onto themselves and turns the integrand into

\frac{tan ^{- 1} ( 1/ t )}{t} = \frac{cot ^{- 1} t}{t}

, so

I = \int_{1/2}^{2} \frac{cot ^{- 1} t}{t} d t

as well. Adding the two forms,

2 I = \int_{1/2}^{2} \frac{tan ^{- 1} x + cot ^{- 1} x}{x} d x = \frac{π}{2} \int_{1/2}^{2} \frac{d x}{x} = \frac{π}{2} (ln 2 - ln \frac{1}{2}) = π ln 2

. Therefore the value of the integral is

\frac{π}{2} ln 2

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