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JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Indefinite Integrals

Let

f (t) = \int (\frac{1 - s i n ( l o g _{e} t )}{1 - c o s ( l o g _{e} t )}) dt, t > 1

.

If

f (e^{π /2}) = - e^{π /2}

and

f (e^{π /4}) = α e^{π /4}

, then

α

equals

(A) $- 1 - 2$ ✓ Correct
(B) $- 1 - 22$
(C) $1 + 2$
(D) $- 1 + 2$

Answer: (A)

Solution

With

x = lo g_{e} t

(so

d t = e^{x} d x

), the integral becomes

\int \frac{1 - sin x}{1 - cos x} e^{x} d x

. Using

1 - cos x = 2 sin^{2} \frac{x}{2}

and

1 - sin x = 1 - 2 sin \frac{x}{2} cos \frac{x}{2}

, the factor is

\frac{1}{2} csc^{2} \frac{x}{2} - cot \frac{x}{2}

. Since

\frac{d}{d x} (- cot \frac{x}{2}) = \frac{1}{2} csc^{2} \frac{x}{2}

, the integrand has the form

(g (x) + g^{'} (x)) e^{x}

with

g (x) = - cot \frac{x}{2}

, so

f (t) = - t cot (\frac{l o g _{e} t}{2}) + C

. The condition

f (e^{π /2}) = - e^{π /2} cot \frac{π}{4} + C = - e^{π /2}

gives

C = 0

. Then

f (e^{π /4}) = - e^{π /4} cot \frac{π}{8} = - e^{π /4} (1 + 2)

, so

α = - 1 - 2

.

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Single correct option

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