Integral Calculus – Indefinite Integration – JEE Main 24 Jan 2026 Shift 1

Solution:

Let $x = \log_e t \Rightarrow t = e^x \Rightarrow dt = e^x dx$.
$$I = \int \frac{1 – \sin x}{1 – \cos x} e^x dx$$
Using half-angle formulas:
$$1 – \cos x = 2\sin^2 \frac{x}{2}$$
$$1 – \sin x = 1 – 2\sin \frac{x}{2} \cos \frac{x}{2}$$
$$I = \int e^x \left[ \frac{1}{2\sin^2 \frac{x}{2}} – \frac{2\sin \frac{x}{2} \cos \frac{x}{2}}{2\sin^2 \frac{x}{2}} \right] dx$$
$$I = \int e^x \left[ \frac{1}{2} cosec^2 \frac{x}{2} – \cot \frac{x}{2} \right] dx$$

This is of the form $\int e^x (f(x) + f'(x)) dx = e^x f(x)$.
Let $f(x) = -\cot \frac{x}{2}$.
Then $f'(x) = -(-cosec^2^2 \frac{x}{2}) \cdot \frac{1}{2} = \frac{1}{2} cosec^2^2 \frac{x}{2}$.
Thus, the integral is:
$$f(t) = -e^x \cot \frac{x}{2} + C = -t \cot\left(\frac{\log_e t}{2}\right) + C$$

Given condition: $f(e^{\pi/2}) = -e^{\pi/2}$.
Substitute $t = e^{\pi/2}$ (so $\log_e t = \pi/2$):
$$-e^{\pi/2} \cot\left(\frac{\pi}{4}\right) + C = -e^{\pi/2}$$
$$-e^{\pi/2}(1) + C = -e^{\pi/2} \Rightarrow C = 0$$
So, $f(t) = -t \cot\left(\frac{\log_e t}{2}\right)$.

Now find $f(e^{\pi/4})$:
Substitute $t = e^{\pi/4}$ (so $\log_e t = \pi/4$):
$$f(e^{\pi/4}) = -e^{\pi/4} \cot\left(\frac{\pi}{8}\right)$$
We know $\cot \frac{\pi}{8} = \sqrt{2} + 1$.
$$f(e^{\pi/4}) = -e^{\pi/4} (\sqrt{2} + 1)$$
Given $f(e^{\pi/4}) = \alpha e^{\pi/4}$:
$$\alpha e^{\pi/4} = -(\sqrt{2} + 1) e^{\pi/4}$$
$$\alpha = -1 – \sqrt{2}$$

Ans. (1)