Q39 / 90 · JEE Main 2019 (09 Apr Shift 2) · Mathematics

If ∫esecxsecxtanxfx+secxtanx+sec2xdx=esecxfx+C, then a possible choice of fx is:

  • A
    secx-tanx-12
  • Correct answer
    B
    secx+tanx+12
  • C
    xsecx+tanx+12
  • D
    secx+xtanx-12

Solution

∫esec⁡x(sec⁡xtan⁡xfx+sec⁡xtan⁡x+sec2⁡xdx=esec⁡xfx+C

Differentiating both sides w.r.t ‘x’ we get

esecxsecxtanxfx+secxtanx+sec2x=esecx⋅secxtanxf(x)+esecxf'x

⇒f'x=sec2x+tanxsecx

⇒fx=∫sec2x+tanxsecxdx

⇒fx=tanx+secx+c, c∈R
Hence, possible choice is fx=secx+tanx+12.

Hence, the correct answer is option: 'B'.

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