Question

Evaluate the integral: \[ \int \frac{1}{9\cos^2 x - 24 \sin x \cos x + 16 \sin^2 x} \, dx = \]

• A. \(\frac{\cos x}{4(3\cos x - 4\sin x)} + c\)
• B. \(\frac{\sin x}{4(3\cos x - 4\sin x)} + c\)
• C. \(\frac{\cos x}{3\cos x - 4\sin x} + c\)
• D. \(\frac{\sin x}{3\cos x - 4\sin x} + c\)

Hint:

When facing trigonometric quadratics in the denominator, try expressing them as a perfect square of linear combinations of sine and cosine.

Answer 1

The Correct Option is A

We are given:
\[ \int \frac{1}{9\cos^2 x - 24 \sin x \cos x + 16 \sin^2 x} \, dx \] Step 1: Recognize the denominator as a perfect square
Try expressing the quadratic form in trigonometric terms as a perfect square:
\[ 9\cos^2 x - 24 \sin x \cos x + 16 \sin^2 x = (3\cos x - 4\sin x)^2 \] Step 2: Substitute the simplified form
Now the integral becomes:
\[ \int \frac{1}{(3\cos x - 4\sin x)^2} \, dx \] Let \(u = 3\cos x - 4\sin x\), then
\[ \frac{du}{dx} = -3\sin x - 4\cos x \Rightarrow \text{Not easily integrable directly, so try reverse engineering} \] We instead test the derivative of:
\[ I = \frac{\cos x}{4(3\cos x - 4\sin x)} \] Using quotient rule:
\[ \frac{dI}{dx} = \frac{-\sin x \cdot 4(3\cos x - 4\sin x) - \cos x \cdot 4(-3\sin x - 4\cos x)}{[4(3\cos x - 4\sin x)]^2} \] After simplifying the numerator, you’ll find:
\[ = \frac{1}{(3\cos x - 4\sin x)^2} \Rightarrow \text{Matches the integrand} \] Step 3: Final Answer
\[ \boxed{\frac{\cos x}{4(3\cos x - 4\sin x)} + c} \]