JEE Main 2026 Jan 24 Shift 1, Mathematics Q11: Methods of Integration

JEE Main2026Jan 24, Shift 1Mathematics

Let $f (t) = \int \frac{1 - sin ( lo g _{e} t )}{1 - cos ( lo g _{e} t )} d t$ , $t > 1$ .

If $f (e^{π /2}) = - e^{π /2}$ and $f (e^{π /4}) = α e^{π /4}$ , then $α$ equals

Solution

Substitute $u = ln t$ , so $t = e^{u}$ and $d t = e^{u} d u$ :

$f = \int \frac{1 - sin u}{1 - cos u} e^{u} d u$ .

Split using $1 - cos u = 2 sin^{2} (u /2)$ and $sin u = 2 sin (u /2) cos (u /2)$ :

$\frac{1 - sin u}{1 - cos u} = \frac{1}{2} csc^{2} (u /2) - cot (u /2)$ .

Let $g (u) = - cot (u /2)$ . Then $g^{'} (u) = \frac{1}{2} csc^{2} (u /2)$ , so the integrand is $[g (u) + g^{'} (u)] e^{u}$ .

Using $\int [g (u) + g^{'} (u)] e^{u} d u = g (u) e^{u} + C$ :

$f = - e^{u} cot (u /2) + C = - t cot (\frac{ln t}{2}) + C$ .

Determine $C$ : $f (e^{π /2}) = - e^{π /2} cot (π /4) + C = - e^{π /2} + C = - e^{π /2} \Rightarrow C = 0$ .

Evaluate at $t = e^{π /4}$ : $f (e^{π /4}) = - e^{π /4} cot (π /8)$ . Using $cot (π /8) = 2 + 1$ :

$f (e^{π /4}) = - (1 + 2) e^{π /4} \Rightarrow α = - 1 - 2$ .

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.