If $\int \frac{dx}{p^2 \sin^2 x + r^2 \cos^2 x} = \frac{1}{12}\tan^{-1}(3\tan x) + c$, then the value of $p\sin x + r\cos x$ can be:
Step-by-Step Solution
Key Concept: The integral form directly reveals the coefficients $p$ and $r$ through comparison, constraining the expression $p\sin x + r\cos x$ to a specific bounded range determined by $\sqrt{p^2 + r^2}$.
We start by evaluating the integral $\int \frac{dx}{p^2 \sin^2 x + r^2 \cos^2 x}$. Dividing numerator and denominator by $\cos^2 x$: $\int \frac{\sec^2 x\,dx}{p^2 \tan^2 x + r^2}$. Let $u = \tan x$, then $\int \frac{du}{p^2 u^2 + r^2} = \frac{1}{pr}\tan^{-1}\left(\frac{pu}{r}\right) + c = \frac{1}{pr}\tan^{-1}\left(\frac{p\tan x}{r}\right) + c$. Comparing with $\frac{1}{12}\tan^{-1}(3\tan x) + c$, we need $\frac{1}{pr} = \frac{1}{12}$ and $\frac{p}{r} = 3$. From these: $pr = 12$ and $p = 3r$, giving $3r^2 = 12$, so $r = 2$ and $p = 6$. Therefore $p\sin x + r\cos x = 6\sin x + 2\cos x$ has range $[-\sqrt{36+4}, \sqrt{36+4}] = [-2\sqrt{10}, 2\sqrt{10}] \approx [-6.32, 6.32]$. Checking options: $\frac{6}{\sqrt{5}} \approx 2.68$ ✓, $\sqrt{5} \approx 2.24$ ✓, $6\sqrt{3} \approx 10.39$ ✗, and $-4$ ✓ all lie within or equal the valid range.
Correct Answer: 1,2,4