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JEE Main 202621 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2026 · 21 January · Evening Shift · Mathematics · Definite Integration

If ∫_ 0 ^ 1 4 ^ -1 (1-2 x+4 x^ 2 ) d x=atan^ -1 (2)-blog_ e (5) ,...

If ∫014cot⁡−1(1−2x+4x2)dx=atan−1(2)−bloge(5)\int_{0}^{1} 4 \cot ^{-1}\left(1-2 x+4 x^{2}\right) \mathrm{d} x=\mathrm{atan}^{-1}(2)-\mathrm{blog}_{\mathrm{e}}(5), where a,b∈N\mathrm{a}, \mathrm{b} \in \mathbf{N}, then (2a+b)(2 \mathrm{a}+\mathrm{b}) is equal to ____\_\_\_\_.

Correct answer

0

Step-by-step solution

Let I=∫01cot⁡−1(1−2x+4x2)dxI=\int_0^1 \cot ^{-1}\left(1-2 x+4 x^2\right) d x
I=∫01(cot⁡−1(2x−1)−cot⁡−1(2x))dxI=\int_0^1\left(\cot ^{-1}(2 x-1)-\cot ^{-1}(2 x)\right) d x

Applying king
I=∫01(−cot⁡−1(2x−1)+cot⁡−1(2x−2))dxI=\int_0^1\left(-\cot ^{-1}(2 x-1)+\cot ^{-1}(2 x-2)\right) d x

From (1) and (2)

2I=∫01(cot⁡−1(2x−2)−cot⁡−1(2x))dx2 I=\int_0^1\left(\cot ^{-1}(2 x-2)-\cot ^{-1}(2 x)\right) d x
=∫01cot⁡−1(2x−2)dx−∫01cot⁡−1(2x)dx=\int_0^1 \cot ^{-1}(2 x-2) d x-\int_0^1 \cot ^{-1}(2 x) d x

Applying King

=∫01cot⁡−1(−2x)dx−∫01cot⁡−1(2x)dx=\int_0^1 \cot ^{-1}(-2 x) d x-\int_0^1 \cot ^{-1}(2 x) d x
=∫01(π−cot⁡−1(2x))dx−∫01cot⁡−1(2x)dx=\int_0^1\left(\pi-\cot ^{-1}(2 x)\right) d x-\int_0^1 \cot ^{-1}(2 x) d x
=∫01(π−2cot⁡−1(2x))dx=\int_0^1\left(\pi-2 \cot ^{-1}(2 x)\right) d x
=π−2∫01(cot⁡−12x)⋅1dx=\pi-2 \int_0^1\left(\cot ^{-1} 2 x\right) \cdot 1 d x

By parts
=π−2[(xcot⁡−12x)01+∫012x1+4x2dx]=\pi-2\left[\left(x \cot ^{-1} 2 x\right)_0^1+\int_0^1 \frac{2 x}{1+4 x^2} d x\right]

Let 1+4x2=t1+4 \mathrm{x}^2=\mathrm{t}
8xdx=dt8 \mathrm{xdx}=\mathrm{dt}
=π−2[cot⁡−12+14∫15dtt]=\pi-2\left[\cot ^{-1} 2+\frac{1}{4} \int_1^5 \frac{\mathrm{dt}}{\mathrm{t}}\right]
=π−2cot⁡−12−12ℓn5=\pi-2 \cot ^{-1} 2-\frac{1}{2} \ell \mathrm{n} 5
22\mathrm{I}=2tan⁡−12−12ℓn5=2 \tan ^{-1} 2-\frac{1}{2} \ell \mathrm{n} 5
⇒4\Rightarrow 4\mathrm{I}=4tan⁡−12−ℓn5=4 \tan ^{-1} 2-\ell \mathrm{n} 5
∴2a+b=8+1=9\therefore 2 \mathrm{a}+\mathrm{b}=8+1=9
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