JEE Main 2026 — 22 January, Evening Shift — Question 6

$\int \frac{\cos y}{1+2 \sin y} d y=\int \frac{d x}{16(\sqrt{9 \sqrt{x}+x})(4+\sqrt{9+\sqrt{x}})}$

$$\begin{aligned} & 4+\sqrt{9+\sqrt{x}}=t \end{aligned}$$

$\frac{1}{2 \sqrt{9+\sqrt{x}}} \times \frac{d x}{2 \sqrt{x}}=1 d x$

$\frac{1}{2} \ell \mathrm{n}|1+2 \sin \mathrm{y}|=\int \frac{4 \mathrm{dt}}{16 \mathrm{t}}+\mathrm{C}$

$\left.\frac{1}{2} \ell \mathrm{n}|1+2 \sin \mathrm{y}|=\frac{1}{4} \ell \mathrm{n} \right\rvert\, 4+\sqrt{9+\sqrt{\mathrm{x}}}+\mathrm{C}$

$\frac{1}{2} \ell \mathrm{n}(2 \sin \mathrm{y}+1)=\frac{1}{4} \ell \mathrm{n}|4+\sqrt{9+\sqrt{\mathrm{x}}}|+\mathrm{C}$

Substituting $\left(256, \frac{\pi}{2}\right)$

$\frac{1}{2} \ell \mathrm{n} 3=\frac{1}{2} \ell \mathrm{n} 3+\mathrm{C} \quad \mathrm{C}=0$

Substituting $(49, \alpha)$

$\frac{1}{2} \ell \mathrm{n}(2 \sin \alpha+1)=\frac{1}{4} \ell \mathrm{n} 8$

$\ell \mathrm{n}(2 \sin \alpha+1)=\frac{1}{2} \ell \mathrm{n} 8$

$\ell \mathrm{n}(2 \sin \alpha+1)=\ell \mathrm{n} 2 \sqrt{2}$

$2 \sin \alpha+1=2 \sqrt{2}$

$2 \sin \alpha=2 \sqrt{2}-1$