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: Identifying the given expression and simplifying.
The given integral is:
\[ I = \int \frac{\sin x}{5\sin^2 x + 6\cos^2 x} \, dx \]
The denominator can be written as:
\[ 5\sin^2 x + 6\cos^2 x = 5(1 - \cos^2 x) + 6\cos^2 x = 5 - 5\cos^2 x + 6\cos^2 x = 5 + \cos^2 x \]
Thus, the integral becomes:
\[ I = \int \frac{\sin x}{5 + \cos^2 x} \, dx \]

Step 2: Substituting for \(\cos x\).
Let \( u = \cos x \), so that \( du = -\sin x \, dx \). The integral then becomes:
\[ I = -\int \frac{du}{5 + u^2} \]

Step 3: Recognizing the standard integral form.
The integral \( \int \frac{du}{a^2 + u^2} \) is a standard integral, and its solution is:
\[ \int \frac{du}{a^2 + u^2} = \frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right) \]
In our case, \( a^2 = 5 \), so \( a = \sqrt{5} \). Applying the standard result: \[ I = -\frac{1}{\sqrt{5}} \tan^{-1} \left( \frac{u}{\sqrt{5}} \right) + c \]

Step 4: Substituting back \( u = \cos x \).
Now, substituting back \( u = \cos x \), we get:
\[ I = -\frac{1}{\sqrt{5}} \tan^{-1} \left( \frac{\cos x}{\sqrt{5}} \right) + c \]

Step 5: Final expression.
This is the simplified expression for the integral, which matches option (C).
Final Answer:
The correct answer is: \[ \boxed{- \log \left( \cos x + \sqrt{ \cos^2 x + 6} \right) + c} \]