Question

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

• A. $\sqrt{40}$
• B. $\sqrt{39}$
• C. $\sqrt{42}$
• D. $\sqrt{41}$

Answer 1

The Correct Option is A

\[\int \frac{\sec^2 x \, dx}{a^2 \tan^2 x + b^2}\]
Let \( \tan x = t \), then \( \sec^2 x \, dx = dt \).
\[= \int \frac{dt}{a^2 t^2 + b^2}\]
\[= \frac{1}{a^2} \int \frac{dt}{t^2 + \left( \frac{b}{a} \right)^2}\]
\[= \frac{1}{ab} \tan^{-1} \left( \frac{ta}{b} \right) + c\]
\[= \frac{1}{ab} \tan^{-1} \left( \frac{a}{b} \tan x \right) + c\]
On comparing, \( \frac{a}{b} = 3 \).
\[ab = 12\]
\[a = 6, \quad b = 2\]
Maximum Value:
The maximum value of \( 6 \sin x + 2 \cos x \) is \( \sqrt{40} \).

Answer 2

To find the maximum value of \(a \sin x + b \cos x\), we know from trigonometry that the expression \(a \sin x + b \cos x\) can have a maximum value given by the amplitude formula for sinusoidal expressions.

The maximum value of \(a \sin x + b \cos x\) is given by:

\(R = \sqrt{a^2 + b^2}\)

In the given problem, we have an expression to compute the integral:

\(\int \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} \, dx = \frac{1}{12} \tan^{-1}(3 \tan x) + \text{constant}\)

The integral given leads to functions involving trigonometric identities. By the structure of the integral, we can infer the presence of a common amplitude term factoring into it.

Analyzing the integral solution structure, particularly the term \(3 \tan x\) in the arc tangent expression, will give us insight about the choice of \(a\) and \(b\):

• By comparing it with known forms, it can be derived that: \(a^2 = 9\) and \(b^2 = 4\) (as \(a^2, b^2\) typically relate to constants in the identity or scale factor for expressions, comparable to the standard circle \(\sin^2 x + \cos^2 x = 1\)).

Thus:

\(a^2 = 36, \, b^2 = 4\)

\(R = \sqrt{36 + 4} = \sqrt{40}\)

This confirms the maximum value of \(a \sin x + b \cos x\) is \(\sqrt{40}\).

Therefore, the correct answer is: \(\sqrt{40}\)