Correct option is (1) \(\sqrt{40}\)
\(\int \frac{\sec ^{2} x d x}{a^{2} \tan ^{2} x+b^{2}}\)
Let \(\tan x=t\)
\(\sec ^{2} d x=d t\)
\(\int \frac{d t}{a^{2} t^{2}+b^{2}}\)
\(\frac{1}{\mathrm{a}^{2}} \int \frac{\mathrm{dt}}{\mathrm{t}^{2}+\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{2}}\)
\(\frac{1}{\mathrm{a}^{2}} \frac{1}{\mathrm{\frac{b}{a}}} \tan ^{-1}\left(\frac{\mathrm{t}}{\mathrm{b}} \mathrm{a}\right)+\mathrm{c}\)
\(\frac{1}{\mathrm{ab}} \tan ^{-1}\left(\frac{\alpha}{\mathrm{b}} \tan \mathrm{x}\right)+\mathrm{c}\)
on comparing \(\frac{\mathrm{a}}{\mathrm{b}}=3\)
ab = 12
a = 6, b = 2
maximum value of
\(6 \sin x+2 \cos x\, is \sqrt{40}\)
