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JEE Main 201816 AprilOnline
Exam
JEE Main
Year
2018
Date
16 April
Mode
Online
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2018 · 16 April · Online · Mathematics · Indefinite Integration

If ∫ x 1 + x + ^ 2 x dx = x - K A ^ - 1 ( Ktanx + 1 A ) + C , ( C...

If ∫tan⁡x1+tan⁡x+tan⁡2xdx=x−KAtan⁡−1(Ktanx+1A)+C∫ \dfrac{\tan x}{1 + \tan x + \tan^{2}x}dx = x - \dfrac{K}{\sqrt{A}}\tan^{- 1}\left(\dfrac{Ktanx + 1}{\sqrt{A}}\right) + C, (CC is a constant of integration), then the ordered pair \left(K, A\right) is equal to

Options

  1. A(2,1)\left(2, 1\right)
  2. B(2,3)\left(2, 3\right)
  3. C(−2,1)\left(- 2, 1\right)
  4. D(−2,3)\left(- 2, 3\right)

Correct answer

B. (2,3)\left(2, 3\right)

Step-by-step solution

Let I=∫tan⁡x1+tan⁡x+tan⁡2xdxI = ∫ \dfrac{\tan x}{1 + \tan x + \tan^{2}x}dx

⇒I=∫(1−sec⁡2x1+tan⁡x+tan⁡2x)dx⇒ I = ∫ \left(1 - \dfrac{\sec^{2}x}{1 + \tan x + \tan^{2}x}\right)dx

⇒I=x−∫dt1+t+t2⇒ I = x - ∫ \dfrac{dt}{1 + t + {t}^{2}}

Where, tan⁡x=t⇒sec⁡2xdx=dt\tan x = t ⇒ \sec^{2}xdx = dt

∴I=x−132tan⁡−1(t+1232)+C∴ I = x - \dfrac{1}{\dfrac{\sqrt{3}}{2}}\tan^{- 1}\left(\dfrac{t + \dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right) + C, where CC is the constant of integration.

=x−23tan⁡−1(2tan⁡x+13)+C= x - \dfrac{2}{\sqrt{3}}\tan^{- 1}\left(\dfrac{2\tan x + 1}{\sqrt{3}}\right) + C

∴K=2 and A=3∴ K = 2 \text{ and } A = 3.

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