Question

Evaluate the integral: \(\int \frac{\sin x}{\sin 4x} \, dx\)

• A. \( -\frac{1}{8}\log \left| \frac{1+\sin x}{1-\sin x} \right| + \frac{1}{4\sqrt{2}}\log \left| \frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x} \right| + C \)
• B. \( \frac{1}{8}\log \left| \frac{1+\sin x}{1-\sin x} \right| - \frac{1}{4\sqrt{2}}\log \left| \frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x} \right| + C \)
• C. \( -\frac{1}{8}\log \left| \frac{1+\sin x}{1-\sin x} \right| + \frac{1}{4\sqrt{2}}\log \left| \frac{1-\sqrt{2}\sin x}{1+\sqrt{2}\sin x} \right| + C \)
• D. None of these

Hint:

When you see \( \sin nx \) in the denominator and \( \sin x \) in the numerator, expanding \( \sin nx \) often allows the \( \sin x \) term to cancel, leading to an integral that can be solved via the \( \sin x = t \) substitution.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The integral involves trigonometric functions in the denominator. We should simplify \( \sin 4x \) to find a suitable substitution.

Step 2: Key Formula or Approach:
Use double angle identities:
\( \sin 4x = 2 \sin 2x \cos 2x = 4 \sin x \cos x \cos 2x \).
Substitute \( \cos 2x = 1 - 2\sin^2 x \) and then use substitution \( t = \sin x \).

Step 3: Detailed Explanation:
The integral is:
\[ I = \int \frac{\sin x}{4 \sin x \cos x \cos 2x} \, dx = \frac{1}{4} \int \frac{1}{\cos x \cos 2x} \, dx \]
Multiply the numerator and denominator by \( \cos x \):
\[ I = \frac{1}{4} \int \frac{\cos x}{\cos^2 x \cos 2x} \, dx = \frac{1}{4} \int \frac{\cos x}{(1 - \sin^2 x)(1 - 2\sin^2 x)} \, dx \]
Let \( t = \sin x \), then \( dt = \cos x \, dx \):
\[ I = \frac{1}{4} \int \frac{1}{(1 - t^2)(1 - 2t^2)} \, dt \]
Using partial fractions on \( \frac{1}{(1-t^2)(1-2t^2)} \):
\[ \frac{1}{(1-t^2)(1-2t^2)} = \frac{2}{1-2t^2} - \frac{1}{1-t^2} \]
Integrating both terms:
\[ I = \frac{1}{4} \left[ \int \frac{2}{1-(\sqrt{2}t)^2} \, dt - \int \frac{1}{1-t^2} \, dt \right] \]
\[ I = \frac{1}{4} \left[ \frac{2}{2\sqrt{2}} \log \left| \frac{1+\sqrt{2}t}{1-\sqrt{2}t} \right| - \frac{1}{2} \log \left| \frac{1+t}{1-t} \right| \right] + C \]
\[ I = \frac{1}{4\sqrt{2}} \log \left| \frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x} \right| - \frac{1}{8} \log \left| \frac{1+\sin x}{1-\sin x} \right| + C \]

Step 4: Final Answer:
The result matches option (A).