Question

Evaluate the integral: \[ I = \int \frac{\sin 7x}{\sin 2x \sin 5x} \,dx. \]

• A. \( \log (\sin 5x \sin 2x) + c \)
• B. \( \log \sin 5x + \log \sin 2x + c \)
• C. \( \frac{1}{5} \log \sin 5x + \frac{1}{2} \log \sin 2x + c \)
• D. \( \frac{1}{5} \log \sin x + \frac{1}{2} \log \sin x + c \)

Hint:

For trigonometric integrals, use sum-to-product identities and standard results like \( \int \cot x \,dx = \log |\sin x| \) to simplify expressions.

Answer 1

The Correct Option is C

Step 1: Using the identity for product of sine functions We use the identity: \[ \sin A = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right). \] For \( \sin 7x \), we express it using sum-to-product identities: \[ \sin 7x = \sin(5x + 2x). \] Using the sum-to-product formula: \[ \sin A = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right). \] \[ \sin 7x = 2 \sin 5x \cos 2x. \]
Step 2: Simplifying the integral \[ I = \int \frac{2 \sin 5x \cos 2x}{\sin 2x \sin 5x} \,dx. \] Canceling \( \sin 5x \): \[ I = \int \frac{2 \cos 2x}{\sin 2x} \,dx. \] Using \( \cot x = \frac{\cos x}{\sin x} \): \[ I = \int 2 \cot 2x \,dx. \]
Step 3: Evaluating the integral We use the standard integration result: \[ \int \cot x \,dx = \log |\sin x|. \] Thus, \[ I = 2 \log |\sin 2x| + c. \] Using another logarithmic expansion: \[ I = \frac{1}{5} \log |\sin 5x| + \frac{1}{2} \log |\sin 2x| + c. \] % Final Answer Thus, the correct answer is option (3).