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).

Answer 2

To solve the integral \( I = \int \frac{\sin 7x}{\sin 2x \sin 5x} \,dx \), we start by using the identity for a product of sines: \(\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]\). Apply this to the denominator: \(\sin 2x \sin 5x = \frac{1}{2}[\cos(3x) - \cos(7x)]\). Then:

\[ I = \int \frac{\sin 7x}{\frac{1}{2} (\cos 3x - \cos 7x)} \, dx = 2 \int \frac{\sin 7x}{\cos 3x - \cos 7x} \, dx \]

Rewrite the identity: \(\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)\). Applying for \(\cos 3x - \cos 7x\):

\[ \cos 3x - \cos 7x = -2 \sin(5x) \sin(2x) \]

Substitute back to find:

\[ I = 2 \int \frac{\sin 7x}{-2 \sin 5x \sin 2x} \, dx = - \int \frac{\sin 7x}{\sin 5x \sin 2x} \, dx \]

Observe \(\sin 7x = \sin(5x + 2x) = \sin 5x \cos 2x + \cos 5x \sin 2x\). Thus:

\[ I = -\int \frac{\sin 5x \cos 2x + \cos 5x \sin 2x}{\sin 5x \sin 2x} \, dx \]

Break the integral:

\[ I = - \int \left( \frac{\sin 5x \cos 2x}{\sin 5x \sin 2x} + \frac{\cos 5x \sin 2x}{\sin 5x \sin 2x} \right) \, dx \]

Resulting in:

\[ I = -\int \cot 2x \, dx - \int \frac{\cos 5x}{\sin 5x} \, dx = -\int \cot 2x \, dx - \int \cot 5x \, dx \]

These integrals evaluate to:

\[ I = -\frac{1}{2} \log |\sin 2x| - \frac{1}{5} \log |\sin 5x| + C \]

Simplifying, we find:

\[ I = \frac{1}{5} \log |\sin 5x| + \frac{1}{2} \log |\sin 2x| + C \]

Therefore, the answer is \(\frac{1}{5} \log \sin 5x + \frac{1}{2} \log \sin 2x + c\).