JEE Main 2025 — 29 January, Evening Shift — Question 61

If $\mathrm{e}^{\int \tan \mathrm{dx} x}=\mathrm{e}^{\ln (\sec \mathrm{x})}=\sec \mathrm{x}$

$\therefore \mathbf{y} \cdot \sec x=\int\left\{\frac{2+\sec x}{(1+2 \sec x)^{2}}\right\} \sec x d x$

$=\int \frac{2 \cos x+1}{(\cos x+2)^{2}} d x$ Let $\cos x=\frac{1-t^{2}}{1+t^{2}}$

$=\int \frac{2\left(\frac{1-t^{2}}{1+t^{2}}\right)+1}{\left(\frac{1-t^{2}}{1+t^{2}}+2\right)^{2}} 2 d t$

$=\int \frac{2-2 t^{2}+1+t^{2}}{\left(1-t^{2}+2+2 t^{2}\right)^{2}} \times 2 d t$

$=2 \int \frac{3-t^{2}}{\left(t^{2}+3\right)^{2}} d t$

Let $\mathrm{t}+\frac{3}{\mathrm{t}}=\mathrm{u}$

$\left(1-\frac{3}{\mathrm{t}^{2}}\right) \mathrm{dt}=\mathrm{du}$

$=-2 \int \frac{\mathrm{du}}{\mathrm{u}^{2}}$ $\begin{aligned} & y \cdot(\sec x)=\frac{2}{u}+c \\ & y \cdot \sec x=\frac{2}{t+\frac{3}{t}}+c \end{aligned}$ At $\mathrm{x}=\frac{\pi}{3}, \mathrm{t}=\tan \frac{\mathrm{x}}{2}=\frac{1}{\sqrt{3}}$

2. $\frac{\sqrt{3}}{10}=\frac{2}{\frac{1}{\sqrt{3}}+3 \sqrt{3}}+c$

3. $\frac{\sqrt{3}}{10}=\frac{2 \sqrt{3}}{10}+\mathrm{c} \Rightarrow \mathrm{C}=0$

At $x=\frac{\pi}{4}, t=\tan \frac{x}{2}=\sqrt{2}-1$

$\therefore \mathrm{y} \cdot \sqrt{2}=\frac{2}{\sqrt{2}-1+\frac{3}{\sqrt{2}-1}}$

y. $\sqrt{2}=\frac{2(\sqrt{2}-1)}{6-2 \sqrt{2}}$

$y=\frac{\sqrt{2}(\sqrt{2}-1)}{2(3-\sqrt{2})}=\frac{1}{\sqrt{2}} \times \frac{2 \sqrt{2}-1}{7}$

$=\frac{4-\sqrt{2}}{14}$