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JEE Main 201912 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2019 · 12 January · Morning Shift · Mathematics · Indefinite Integration

The integral ∫ ( x)dx , is equal to

The integral ∫cos⁡(ln⁡x)dx∫ \cos \left(\ln x\right)dx, is equal to

Options

  1. Ax2(cos⁡(ln⁡x)−sin⁡(ln⁡x))+C\dfrac{x}{2}\left(\cos \left(\ln x\right) - \sin \left(\ln x\right)\right) + C
  2. Bx(cos⁡(ln⁡x)−sin⁡(ln⁡x))+Cx\left(\cos \left(\ln x\right) - \sin \left(\ln x\right)\right) + C
  3. Cx(cos⁡(ln⁡x)+sin⁡(ln⁡x))+Cx\left(\cos \left(\ln x\right) + \sin \left(\ln x\right)\right) + C
  4. Dx2(cos⁡(ln⁡x)+sin⁡(ln⁡x))+C\dfrac{x}{2}\left(\cos \left(\ln x\right) + \sin \left(\ln x\right)\right) + C

Correct answer

D. x2(cos⁡(ln⁡x)+sin⁡(ln⁡x))+C\dfrac{x}{2}\left(\cos \left(\ln x\right) + \sin \left(\ln x\right)\right) + C

Step-by-step solution

Given, I=∫cos⁡(ln⁡x)dxI = ∫ \cos \left(\ln x\right)dx

Let, ln⁡x=t⇒x=et⇒dx=etdt\ln x = t ⇒ x = {e}^{t} ⇒ dx = {e}^{t}dt

∴I=∫etcostdt∴ I = ∫ {e}^{t}costdt

∫etcostdt=12∫et((cos⁡t+sin⁡t)+(cos⁡t−sin⁡t))dt∫ {e}^{t}costdt = \dfrac{1}{2} ∫ {e}^{t}\left(\left(\cos t + \sin t\right) + \left(\cos t - \sin t\right)\right)dt

∫etcostdt=12et(cos⁡t+sin⁡t)+C∫ {e}^{t}costdt = \dfrac{1}{2}{e}^{t}\left(\cos t + \sin t\right) + C, where CC is the constant of integration.

=x2(cos⁡(ln⁡x)+sin⁡(ln⁡x))+C= \dfrac{x}{2}\left(\cos \left(\ln x\right) + \sin \left(\ln x\right)\right) + C.

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