Question

Evaluate the integral: \[ \int \frac{\sin x}{5\sin^2 x + 6\cos^2 x} \, dx \]

• A. \( \log \left( \cos x + \sqrt{ \cos^2 x + 5} \right) + c \)
• B. \( \log \left( \sin x + \sqrt{6 \cos^2 x + 5} \right) + c \)
• C. \( - \log \left( \cos x + \sqrt{ \cos^2 x + 6} \right) + c \)
• D. \( - \log \left( \cos x + \sqrt{ \cos^2 x + 5} \right) + c \)

Hint:

Remember to simplify the denominator and use standard integral results for inverse trigonometric functions.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The integral features \( \sin x \) in the numerator and a square root involving trig squares in the denominator. Converting everything in the denominator to \( \cos x \) allows for a simple substitution.
Step 2: Key Formula or Approach:
Use \( \sin^2 x = 1 - \cos^2 x \) and then substitute \( \cos x = u \).
Step 3: Detailed Explanation:
Denominator: \( \sqrt{5(1 - \cos^2 x) + 6\cos^2 x} = \sqrt{5 - 5\cos^2 x + 6\cos^2 x} = \sqrt{\cos^2 x + 5} \).
Integral becomes:
\[ \int \frac{\sin x}{\sqrt{\cos^2 x + 5}} \text{ d}x \] Let \( \cos x = u \Rightarrow -\sin x \text{ d}x = \text{d}u \Rightarrow \sin x \text{ d}x = -\text{d}u \).
\[ \text{I} = \int \frac{-\text{d}u}{\sqrt{u^2 + 5}} = -\int \frac{\text{d}u}{\sqrt{u^2 + (\sqrt{5})^2}} \] Using the standard formula \( \int \frac{\text{d}x}{\sqrt{x^2 + a^2}} = \ln|x + \sqrt{x^2 + a^2}| \):
\[ \text{I} = -\ln|u + \sqrt{u^2 + 5}| + \text{c} \] Substitute \( u = \cos x \):
\[ \text{I} = -\ln|\cos x + \sqrt{\cos^2 x + 5}| + \text{c} \]
Step 4: Final Answer:
The integral is \( -\log \left( \cos x + \sqrt{\cos^2 x + 5} \right) + \text{c} \).