The value of the integral 1/2^2 ^-1 xx dx is equal to: — Solution | JEE Mathematics

MCQMediumJEE 2023Definite Integrals

The value of the integral $\int_{1/2}^2 \frac{\tan^{-1} x}{x} \, dx$ is equal to:

Answer

Correct answer:D

Standard Method

Given:

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

Find: The value of the integral and the correct option.

Use the substitution

x = \frac{1}{t}, \quad dx = -\frac{1}{t^2} \, dt

The limits change as

x = \frac{1}{2} \Rightarrow t = 2, \quad x = 2 \Rightarrow t = \frac{1}{2}

Using symmetry by reciprocal substitution

After substitution,

I = \int_2^{1/2} \frac{\tan^{-1}\left(\frac{1}{t}\right)}{\frac{1}{t}} \cdot \left(-\frac{1}{t^2}\right) dt

which simplifies to

I = \int_{1/2}^2 \frac{\tan^{-1}\left(\frac{1}{t}\right)}{t} \, dt

Renaming $t$ as $x$ ,

I = \int_{1/2}^2 \frac{\tan^{-1}\left(\frac{1}{x}\right)}{x} \, dx

Add the integral to its transformed form

Add the two expressions for $I$ :

2I = \int_{1/2}^2 \frac{\tan^{-1} x}{x} \, dx + \int_{1/2}^2 \frac{\tan^{-1}\left(\frac{1}{x}\right)}{x} \, dx

So,

2I = \int_{1/2}^2 \frac{\tan^{-1} x + \tan^{-1}\left(\frac{1}{x}\right)}{x} \, dx

For $x > 0$ ,

\tan^{-1} x + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2}

Hence,

2I = \int_{1/2}^2 \frac{\pi/2}{x} \, dx = \frac{\pi}{2} \int_{1/2}^2 \frac{1}{x} \, dx

Now,

2I = \frac{\pi}{2} \left[\log_e x\right]_{1/2}^2

2I = \frac{\pi}{2} \left(\log_e 2 - \log_e \frac{1}{2}\right)

Since

\log_e \frac{1}{2} = -\log_e 2

we get

2I = \frac{\pi}{2} \left(2\log_e 2\right) = \pi \log_e 2

Therefore,

I = \frac{\pi}{2} \log_e 2

So the value of the integral is $\frac{\pi}{2} \log_e 2$ , and the correct option is D.

The solution lists Option A, but the working clearly gives $\frac{\pi}{2} \log_e 2$ , which matches Option D.

Using the identity $\tan^{-1} x + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2}$ without checking that $x > 0$ . This identity depends on the sign of $x$ ; here it is valid because the interval is $\left[\frac{1}{2}, 2\right]$ . Always verify the domain before applying it.
Making an error while changing limits under the substitution $x = \frac{1}{t}$ . If the limits are not updated correctly, the sign of the integral becomes wrong. Always convert both endpoints before simplifying.
Forgetting the negative sign in $dx = -\frac{1}{t^2} \, dt$ . That sign is compensated by the reversal of limits. Track both the differential and the new bounds carefully.

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