Step 1: Understanding the Concept:
To evaluate this integral, we first examine the structure of the integrand.
The numerator contains odd powers of \( \sin x \), and the denominator is \( \cos 2x \).
In such cases, it is usually beneficial to convert as much of the expression as possible into terms of \( \cos x \) and use the substitution \( t = \cos x \).
This works because the leftover \( \sin x \) in the numerator acts as the differential \( -dt \).
Step 2: Key Formula or Approach:
Identity 1: \( \sin^2 x = 1 - \cos^2 x \).
Identity 2: \( \cos 2x = 2\cos^2 x - 1 \).
Substitution: \( t = \cos x \implies dt = -\sin x \, dx \).
Step 3: Detailed Explanation:
Factor out \( \sin x \) from the numerator:
\[ \int \frac{\sin x(1 + \sin^2 x)}{2\cos^2 x - 1} dx \]
Replace \( \sin^2 x \) with \( 1 - \cos^2 x \):
\[ \int \frac{\sin x(1 + 1 - \cos^2 x)}{2\cos^2 x - 1} dx = \int \frac{\sin x(2 - \cos^2 x)}{2\cos^2 x - 1} dx \]
Now substitute \( t = \cos x \), then \( dt = -\sin x \, dx \) or \( \sin x \, dx = -dt \):
\[ I = \int \frac{2 - t^2}{2t^2 - 1} (-dt) = \int \frac{t^2 - 2}{2t^2 - 1} dt \]
Perform polynomial division or algebraic manipulation on the integrand:
\[ \frac{t^2 - 2}{2t^2 - 1} = \frac{1}{2} \left( \frac{2t^2 - 4}{2t^2 - 1} \right) = \frac{1}{2} \left( \frac{2t^2 - 1 - 3}{2t^2 - 1} \right) = \frac{1}{2} \left( 1 - \frac{3}{2t^2 - 1} \right) \]
Now integrate term by term:
\[ I = \frac{1}{2} \int 1 dt - \frac{3}{2} \int \frac{1}{2t^2 - 1} dt \]
The first part is \( \frac{1}{2}t \).
For the second part, factor out 2 from the denominator:
\[ \frac{3}{2} \int \frac{1}{2(t^2 - 1/2)} dt = \frac{3}{4} \int \frac{1}{t^2 - (1/\sqrt{2})^2} dt \]
Using the standard formula \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| \) where \( a = 1/\sqrt{2} \):
\[ \text{Integral} = \frac{3}{4} \cdot \frac{1}{2(1/\sqrt{2})} \log \left| \frac{t - 1/\sqrt{2}}{t + 1/\sqrt{2}} \right| \]
\[ \text{Integral} = \frac{3\sqrt{2}}{8} \log \left| \frac{\sqrt{2}t - 1}{\sqrt{2}t + 1} \right| = \frac{3}{4\sqrt{2}} \log \left| \frac{\sqrt{2}t - 1}{\sqrt{2}t + 1} \right| \]
Combining everything back together:
\[ I = \frac{1}{2} \cos x - \frac{3}{4\sqrt{2}} \log \left| \frac{\sqrt{2}\cos x - 1}{\sqrt{2}\cos x + 1} \right| + c \]
Comparing this result with the given form \( P \cos x + Q \log |...| \):
\( P = 1/2 \) and \( Q = -3/(4\sqrt{2}) \).
Step 4: Final Answer:
The values of P and Q are \( 1/2 \) and \( -3/(4\sqrt{2}) \) respectively.
This matches Option (B).