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MathMedMain 2026 • 4th Apr - Shift 2

Definite Integral of Inverse Cotangent Function

The integral ∫01cot⁡−1(1+x+x2)dx\int_{0}^{1} \cot^{-1}\left(1 + x + x^2\right) dx is equal to:

Options

A

2tan⁡−12+12log⁡e(54)+π22\tan^{-1}2 + \frac{1}{2}\log_e\left(\frac{5}{4}\right) + \frac{\pi}{2}

B

2tan⁡−12+12log⁡e(54)−π22\tan^{-1}2 + \frac{1}{2}\log_e\left(\frac{5}{4}\right) - \frac{\pi}{2}

C

2tan⁡−12−12log⁡e(54)+π22\tan^{-1}2 - \frac{1}{2}\log_e\left(\frac{5}{4}\right) + \frac{\pi}{2}

D

2tan⁡−12−12log⁡e(54)−π22\tan^{-1}2 - \frac{1}{2}\log_e\left(\frac{5}{4}\right) - \frac{\pi}{2}

Correct

Topics & Concepts

IntegralsDefinite IntegrationProperties of Definite IntegralsInverse Trigonometric Functions

Step-by-Step Solution

To evaluate the definite integral I=∫01cot⁡−1(1+x+x2) dxI = \int_{0}^{1} \cot^{-1}\left(1 + x + x^2\right) \, dx

Step 1: Simplify the integrand using inverse trigonometric identities

We know that for positive arguments, cot⁡−1(y)=tan⁡−1(1y)\cot^{-1}(y) = \tan^{-1}\left(\frac{1}{y}\right). Thus: cot⁡−1(1+x+x2)=tan⁡−1(11+x+x2)\cot^{-1}\left(1 + x + x^2\right) = \tan^{-1}\left(\frac{1}{1 + x + x^2}\right)

We can rewrite the fraction inside the tan⁡−1\tan^{-1} function by expressing the numerator as the difference of two terms in the denominator: 11+x+x2=11+x(x+1)=(x+1)−x1+(x+1)x\frac{1}{1 + x + x^2} = \frac{1}{1 + x(x+1)} = \frac{(x+1) - x}{1 + (x+1)x}

Using the standard identity tan⁡−1(a−b1+ab)=tan⁡−1(a)−tan⁡−1(b)\tan^{-1}\left(\frac{a - b}{1 + ab}\right) = \tan^{-1}(a) - \tan^{-1}(b), we get: tan⁡−1((x+1)−x1+(x+1)x)=tan⁡−1(x+1)−tan⁡−1(x)\tan^{-1}\left(\frac{(x+1) - x}{1 + (x+1)x}\right) = \tan^{-1}(x+1) - \tan^{-1}(x)

Step 2: Split the integral into two parts

Now, the integral becomes: I=∫01(tan⁡−1(x+1)−tan⁡−1(x)) dx=∫01tan⁡−1(x+1) dx−∫01tan⁡−1(x) dxI = \int_{0}^{1} \left( \tan^{-1}(x+1) - \tan^{-1}(x) \right) \, dx = \int_{0}^{1} \tan^{-1}(x+1) \, dx - \int_{0}^{1} \tan^{-1}(x) \, dx

For the first integral, substitute u=x+1  ⟹  du=dxu = x+1 \implies du = dx. The limits change from x=0→1x = 0 \to 1 to u=1→2u = 1 \to 2: ∫01tan⁡−1(x+1) dx=∫12tan⁡−1(u) du\int_{0}^{1} \tan^{-1}(x+1) \, dx = \int_{1}^{2} \tan^{-1}(u) \, du

Therefore, the integral II can be written as: I=∫12tan⁡−1(x) dx−∫01tan⁡−1(x) dxI = \int_{1}^{2} \tan^{-1}(x) \, dx - \int_{0}^{1} \tan^{-1}(x) \, dx

Step 3: Evaluate the integrals using integration by parts

The standard indefinite integral of tan⁡−1(x)\tan^{-1}(x) is: ∫tan⁡−1(x) dx=xtan⁡−1(x)−12log⁡e(1+x2)\int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \frac{1}{2} \log_e\left(1 + x^2\right)

  1. Evaluating ∫12tan⁡−1(x) dx\int_{1}^{2} \tan^{-1}(x) \, dx: ∫12tan⁡−1(x) dx=[xtan⁡−1(x)−12log⁡e(1+x2)]12\int_{1}^{2} \tan^{-1}(x) \, dx = \left[ x \tan^{-1}(x) - \frac{1}{2} \log_e\left(1 + x^2\right) \right]_{1}^{2} =(2tan⁡−1(2)−12log⁡e(5))−(1⋅tan⁡−1(1)−12log⁡e(2))= \left( 2 \tan^{-1}(2) - \frac{1}{2} \log_e(5) \right) - \left( 1 \cdot \tan^{-1}(1) - \frac{1}{2} \log_e(2) \right) Since tan⁡−1(1)=π4\tan^{-1}(1) = \frac{\pi}{4}: =2tan⁡−1(2)−12log⁡e(5)−π4+12log⁡e(2)= 2 \tan^{-1}(2) - \frac{1}{2} \log_e(5) - \frac{\pi}{4} + \frac{1}{2} \log_e(2)

  2. Evaluating ∫01tan⁡−1(x) dx\int_{0}^{1} \tan^{-1}(x) \, dx: ∫01tan⁡−1(x) dx=[xtan⁡−1(x)−12log⁡e(1+x2)]01\int_{0}^{1} \tan^{-1}(x) \, dx = \left[ x \tan^{-1}(x) - \frac{1}{2} \log_e\left(1 + x^2\right) \right]_{0}^{1} =(1⋅tan⁡−1(1)−12log⁡e(2))−(0−0)=π4−12log⁡e(2)= \left( 1 \cdot \tan^{-1}(1) - \frac{1}{2} \log_e(2) \right) - (0 - 0) = \frac{\pi}{4} - \frac{1}{2} \log_e(2)

Step 4: Combine the results

Subtract the second value from the first: I=(2tan⁡−1(2)−12log⁡e(5)−π4+12log⁡e(2))−(π4−12log⁡e(2))I = \left( 2 \tan^{-1}(2) - \frac{1}{2} \log_e(5) - \frac{\pi}{4} + \frac{1}{2} \log_e(2) \right) - \left( \frac{\pi}{4} - \frac{1}{2} \log_e(2) \right) =2tan⁡−1(2)−(π4+π4)−12log⁡e(5)+log⁡e(2)= 2 \tan^{-1}(2) - \left( \frac{\pi}{4} + \frac{\pi}{4} \right) - \frac{1}{2} \log_e(5) + \log_e(2) =2tan⁡−1(2)−π2−12(log⁡e(5)−2log⁡e(2))= 2 \tan^{-1}(2) - \frac{\pi}{2} - \frac{1}{2} \left( \log_e(5) - 2 \log_e(2) \right) =2tan⁡−1(2)−π2−12(log⁡e(5)−log⁡e(4))= 2 \tan^{-1}(2) - \frac{\pi}{2} - \frac{1}{2} \left( \log_e(5) - \log_e(4) \right) =2tan⁡−1(2)−12log⁡e(54)−π2= 2 \tan^{-1}(2) - \frac{1}{2} \log_e\left(\frac{5}{4}\right) - \frac{\pi}{2}

Thus, the correct option is D.

Definite Integral of Inverse Cotangent Function | Mathematics PYQ Solution - JEE Challenger