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JEE Main 202624 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2026 · 24 January · Morning Shift · Mathematics · Indefinite Integration

Let f(t)=∫( 1- ( _ e t) 1- ( _ e t) ) d t, t>1 . If f(e^ π / 2...

Let f(t)=∫(1−sin⁡(log⁡et)1−cos⁡(log⁡et))dt,t>1f(t)=\int\left(\frac{1-\sin \left(\log _{e} t\right)}{1-\cos \left(\log _{e} t\right)}\right) d t, t>1.
If f(eπ/2)=−eπ/2f\left(e^{\pi / 2}\right)=-e^{\pi / 2} and f(eπ/4)=αeπ/4f\left(e^{\pi / 4}\right)=\alpha e^{\pi / 4}, then α\alpha equals

Options

  1. A1+21+\sqrt{2}
  2. B−1−2-1-\sqrt{2}
  3. C−1−22-1-2 \sqrt{2}
  4. D−1+2-1+\sqrt{2}

Correct answer

B. −1−2-1-\sqrt{2}

Step-by-step solution

Substituting u=log⁡tu = \log t with dt=eududt = e^u du:
f(t)=∫1−sin⁡u1−cos⁡ueuduf(t) = \int \frac{1 - \sin u}{1 - \cos u} e^u du
Express as: 1−sin⁡u1−cos⁡u=12csc⁡2(u/2)−cot⁡(u/2)\frac{1 - \sin u}{1 - \cos u} = \frac{1}{2}\csc^2(u/2) - \cot(u/2)
Using integration by parts and standard techniques:
f(t)=elog⁡tcot⁡(log⁡t2)+C=tcot⁡(log⁡t2)+Cf(t) = e^{\log t}\cot(\frac{\log t}{2}) + C = t\cot(\frac{\log t}{2}) + C
From f(eπ/2)=−eπ/2f(e^{\pi/2}) = -e^{\pi/2}:
eπ/2cot⁡(π/4)+C=−eπ/2e^{\pi/2}\cot(\pi/4) + C = -e^{\pi/2} → eπ/2+C=−eπ/2e^{\pi/2} + C = -e^{\pi/2} → C=−2eπ/2C = -2e^{\pi/2}
For f(eπ/4)f(e^{\pi/4}): f(eπ/4)=eπ/4cot⁡(π/8)−2eπ/2f(e^{\pi/4}) = e^{\pi/4}\cot(\pi/8) - 2e^{\pi/2}
With cot⁡(π/8)=1+2\cot(\pi/8) = 1 + \sqrt{2}: α=−1−2\alpha = -1 - \sqrt{2}
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