If $\int \frac{1}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x=\frac{1}{12} \tan ^{-1}(3 \tan x)+\text{constant}$, then the maximum value of $a \sin x+b \cos x$, is

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Indefinite Integrals

\int \frac{1}{a ^{2} s i n ^{2} x + b ^{2} c o s ^{2} x} d x = \frac{1}{12} tan^{- 1} (3 tan x) + constant

, then the maximum value of

a sin x + b cos x

, is

Answer: (C)

Given

\int \frac{d x}{a ^{2} sin ^{2} x + b ^{2} cos ^{2} x} = \frac{1}{ab} tan^{- 1} (\frac{a}{b} tan x) + C

. Comparing with

\frac{1}{12} tan^{- 1} (3 tan x)

gives

\frac{a}{b} = 3

and

ab = 12

, so

a = 6, b = 2

. The maximum of

a sin x + b cos x

a^{2} + b^{2} = 36 + 4 = 40

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Single correct option

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