JEE Main 2024 — 6 April, Shift 2 — Question 19

$\int \frac{\sec ^{2} x d x}{a^{2} \tan ^{2} x+b^{2}}$

let $\tan \mathrm{x}=\mathrm{t}$

$\sec ^{2} d x=d t$

$\int \frac{d t}{a^{2} t^{2}+b^{2}}$

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

$\frac{1}{\mathrm{a}^{2}} \frac{1}{\frac{\mathrm{~b}}{\mathrm{a}}} \tan ^{-1}\left(\frac{\mathrm{t}}{\mathrm{b}} \mathrm{a}\right)+\mathrm{c}$

$\frac{1}{\mathrm{ab}} \tan ^{-1}\left(\frac{\alpha}{\mathrm{~b}} \tan \mathrm{x}\right)+\mathrm{c}$ on comparing $\frac{\mathrm{a}}{\mathrm{b}}=3$

$\mathrm{ab}=12$

$a=6, b=2$

maximum value of

$6 \sin x+2 \cos x$ is $\sqrt{40}$