JEE MainMathematics

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Definite Integral of an Inverse Trigonometric Function via Algebraic Decomposition

Q: How do we identify the decomposition of 1 - 2x + 4x^2?

Notice that 1 - 2x + 4x^2 = 1 + 2x(2x-1). In the cotangent difference formula cot(A-B) = (cot A cot B + 1)/(cot B - cot A), letting cot A = 2x and cot B = 2x-1 gives (2x(2x-1)+1)/((2x-1)-2x) = -(1 - 2x + 4x^2), which directly gives cot^(-1)(1-2x+4x^2) = cot^(-1)(2x-1) - cot^(-1)(2x).

If integral from 0 to 1 of (4 cot^(-1)(1 - 2x + 4x^2)) dx = a tan^(-1)(2) - b log_e(5), where a, b ∈ ℕ, then (2a+b) is equal to.

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Correct answer

The value of $(2 a + b)$ is $9$ , where $a = 4$ and $b = 1$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

The integral to evaluate is $J = 4 I = 4 \int_{0}^{1} \cot^{- 1} (1 - 2 x + 4 x^{2}) d x = a a n^{- 1} (2) - b \log_{e} (5)$ where $a, b \in N$ .

02Approach

Rewrite the argument of $\cot^{- 1}$ or convert to $a n^{- 1}$ using the identity $\cot^{-1}(1 - 2x + 4x^2) = an^{-1}\left( rac{2x - (2x-1)}{1 + 2x(2x-1)} ight) = an^{-1}(2x) - an^{-1}(2x-1)$ , or apply the cotangent subtraction identity directly.

03Execute

Let $I = \int_0^1 \cot^{-1}(1 - 2x + 4x^2) \, dx = \int_0^1 \left(\cot^{-1}(2x-1) - \cot^{-1}(2x) ight) dx$ . Using the property $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ on $I$ , we obtain $I = \int_0^1 \left(\cot^{-1}(1-2x) - \cot^{-1}(2-2x) ight) dx$ . Since $\cot^{- 1} (- y) = π - \cot^{- 1} (y)$ , adding the two representations or simplifying gives $2I = \int_0^1 \left(\pi - 2\cot^{-1}(2x) ight) dx = \pi - 2 \int_0^1 \cot^{-1}(2x) \, dx$ .

04Execute

Evaluate $\int_{0}^{1} \cot^{- 1} (2 x) d x$ using integration by parts: $\int_0^1 1 \cdot \cot^{-1}(2x) \, dx = \left[ x \cot^{-1}(2x) ight]_0^1 - \int_0^1 x \left(- rac{2}{1 + 4x^2} ight) dx = \cot^{-1}(2) + rac{1}{4} \int_0^1 rac{8x \, dx}{1+4x^2} = \cot^{-1}(2) + rac{1}{4} \ln(5)$ .

05Execute

Substitute back into $2 I$ : $2I = \pi - 2\left(\cot^{-1}(2) + rac{1}{4}\ln 5 ight) = (\pi - 2\cot^{-1}(2)) - rac{1}{2}\ln 5 = 2\left( rac{\pi}{2} - \cot^{-1}(2) ight) - rac{1}{2}\ln 5 = 2 an^{-1}(2) - rac{1}{2}\ln 5$ . Therefore, $J = 4 I = 2 (2 I) = 4 a n^{- 1} (2) - \ln (5)$ .

06Execute

Comparing with $a a n^{- 1} (2) - b \log_{e} (5)$ gives $a = 4$ and $b = 1$ . Thus, $2 a + b = 2 (4) + 1 = 9$ .

✓Verify

Check that $a, b \in N$ : $a = 4 \in N$ and $b = 1 \in N$ . Also, since $1 - 2 x + 4 x^{2} = (2 x - 1 / 2)^{2} + 3 / 4 > 0$ for all $x \in [0, 1]$ , the inverse cotangent identity $\cot^{- 1} (u) = a n^{- 1} (1 / u)$ holds without any branch-shift issues.

Hints that build this answer step by step

How can the integrand $\cot^{- 1} (1 - 2 x + 4 x^{2})$ be rewritten using the identity for the difference of inverse tangents?
$\tan^{- 1} (2 x) + \tan^{- 1} (1 - 2 x)$
Using the symmetry property $\int_{0}^{1} f (x) d x = \int_{0}^{1} f (1 - x) d x$ or direct substitution, what is $\int_{0}^{1} \tan^{- 1} (1 - 2 x) d x$ ?
$0$
What is the value of the remaining integral $4 \int_{0}^{1} \tan^{- 1} (2 x) d x$ evaluated using integration by parts?
$4 \tan^{- 1} (2) - \log_{e} (5)$
Comparing $4 \tan^{- 1} (2) - \log_{e} (5)$ with $a \tan^{- 1} (2) - b \log_{e} (5)$ , what is the final value of $2 a + b$ ?
$9$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

How do we identify the decomposition of

1 - 2 x + 4 x^{2}

Notice that $1 - 2 x + 4 x^{2} = 1 + 2 x (2 x - 1)$ . In the cotangent difference formula $\cot (A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A}$ , letting $\cot A = 2 x$ and $\cot B = 2 x - 1$ gives $\frac{2 x (2 x - 1) + 1}{(2 x - 1) - 2 x} = - (1 - 2 x + 4 x^{2})$ , which directly gives $\cot^{- 1} (1 - 2 x + 4 x^{2}) = \cot^{- 1} (2 x - 1) - \cot^{- 1} (2 x)$ .

Answer

Accepted answer: 9

The value of $(2 a + b)$ is $9$ , where $a = 4$ and $b = 1$ .

Why each option works or fails

Step-by-step solution

Given: The integral to evaluate is $J = 4 I = 4 \int_{0}^{1} \cot^{- 1} (1 - 2 x + 4 x^{2}) d x = a a n^{- 1} (2) - b \log_{e} (5)$ where $a, b \in N$ .
Approach: Rewrite the argument of $\cot^{- 1}$ or convert to $a n^{- 1}$ using the identity $\cot^{-1}(1 - 2x + 4x^2) = an^{-1}\left( rac{2x - (2x-1)}{1 + 2x(2x-1)} ight) = an^{-1}(2x) - an^{-1}(2x-1)$ , or apply the cotangent subtraction identity directly.
Execute: Let $I = \int_0^1 \cot^{-1}(1 - 2x + 4x^2) \, dx = \int_0^1 \left(\cot^{-1}(2x-1) - \cot^{-1}(2x) ight) dx$ . Using the property $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ on $I$ , we obtain $I = \int_0^1 \left(\cot^{-1}(1-2x) - \cot^{-1}(2-2x) ight) dx$ . Since $\cot^{- 1} (- y) = π - \cot^{- 1} (y)$ , adding the two representations or simplifying gives $(2 a + b)$ .
Execute: Evaluate $\int_{0}^{1} \cot^{- 1} (2 x) d x$ using integration by parts: $\int_0^1 1 \cdot \cot^{-1}(2x) \, dx = \left[ x \cot^{-1}(2x) ight]_0^1 - \int_0^1 x \left(- rac{2}{1 + 4x^2} ight) dx = \cot^{-1}(2) + rac{1}{4} \int_0^1 rac{8x \, dx}{1+4x^2} = \cot^{-1}(2) + rac{1}{4} \ln(5)$ .
Execute: Substitute back into $2 I$ : $2I = \pi - 2\left(\cot^{-1}(2) + rac{1}{4}\ln 5 ight) = (\pi - 2\cot^{-1}(2)) - rac{1}{2}\ln 5 = 2\left( rac{\pi}{2} - \cot^{-1}(2) ight) - rac{1}{2}\ln 5 = 2 an^{-1}(2) - rac{1}{2}\ln 5$ . Therefore, $J = 4 I = 2 (2 I) = 4 a n^{- 1} (2) - \ln (5)$ .
Execute: Comparing with $a a n^{- 1} (2) - b \log_{e} (5)$ gives $a = 4$ and $b = 1$ . Thus, $2 a + b = 2 (4) + 1 = 9$ .
Verify: Check that $a, b \in N$ : $a = 4 \in N$ and $b = 1 \in N$ . Also, since $1 - 2 x + 4 x^{2} = (2 x - 1 / 2)^{2} + 3 / 4 > 0$ for all $x \in [0, 1]$ , the inverse cotangent identity $\cot^{- 1} (u) = a n^{- 1} (1 / u)$ holds without any branch-shift issues.

Definite Integral of an Inverse Trigonometric Function via Algebraic Decomposition

What feels right?

The value of $(2 a + b)$ is $9$ , where $a = 4$ and $b = 1$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

05Execute

06Execute

✓Verify

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Definite Integral of an Inverse Trigonometric Function via Algebraic Decomposition

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