If 1 a 2 sin 2 x + b 2 cos 2 x d x = 1 12 tan - 1 ( 3 tan x ) + constant , then the maximum value of a sin x + b cos x is: [2024]

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Solution

Q.
$If \int \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x = \frac{1}{12} \tan^{- 1} (3 \tan x) + constant,$ $then the maximum value of a \sin x + b \cos x is:$ [2024]

1 $\sqrt{41}$

2 $\sqrt{42}$

3 $\sqrt{40}$

4 $\sqrt{39}$

Ans.
(3)

We have, $\int \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x = \frac{1}{12} \tan^{- 1} (3 \tan x) + c$

Let $I = \int \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x = \int \frac{\sec^{2} x}{a^{2} \tan^{2} x + b^{2}} d x$

Let $\tan x = t \Rightarrow \sec^{2} x d x = d t$

$∴$ $I = \int \frac{d t}{{(a t)}^{2} + b^{2}}$

$= \frac{1}{b a} \tan^{- 1} (\frac{a t}{b}) + c = \frac{1}{b a} \tan^{- 1} (\frac{a}{b} \tan x) + c$

$\Rightarrow a b = 12 and \frac{a}{b} = 3$

$\Rightarrow a^{2} = 36 and b^{2} = 4$

Now, $- \sqrt{a^{2} + b^{2}} \leq a \sin x + b \cos x \leq \sqrt{a^{2} + b^{2}}$

Required maximum value $= \sqrt{36 + 4} = \sqrt{40}$

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