Step 1: Understanding the Concept:
The given problem is an indefinite integral involving trigonometric functions.
To solve this, we must simplify the integrand into a form that allows for a standard substitution.
The presence of \( \sin x \) and its higher powers in the numerator suggests that substituting \( u = \cos x \) would be effective, as the derivative \( -\sin x \) appears naturally.
The denominator \( \cos 2x \) is a double-angle term that must be converted to an expression involving \( \cos x \) to maintain consistency with our proposed substitution.
Key Formula or Approach:
1. Trigonometric Identity: \( \sin^2 x = 1 - \cos^2 x \).
2. Double Angle Formula: \( \cos 2x = 2 \cos^2 x - 1 \).
3. Standard Integral: \( \int \frac{1}{x^2 - a^2} dx = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C \).
Step 2: Detailed Explanation:
Let us start by rewriting the integral \( I \):
\[ I = \int \frac{\sin x + \sin^3 x}{\cos 2x} dx \]
Factor out \( \sin x \) from the numerator:
\[ I = \int \frac{\sin x (1 + \sin^2 x)}{\cos 2x} dx \]
Now, we apply the identity \( \sin^2 x = 1 - \cos^2 x \) in the numerator and \( \cos 2x = 2 \cos^2 x - 1 \) in the denominator:
\[ I = \int \frac{\sin x (1 + 1 - \cos^2 x)}{2 \cos^2 x - 1} dx = \int \frac{\sin x (2 - \cos^2 x)}{2 \cos^2 x - 1} dx \]
Perform the substitution \( u = \cos x \).
Differentiating both sides with respect to \( x \), we get \( \frac{du}{dx} = -\sin x \), which implies \( \sin x dx = -du \).
Substitute these into the integral:
\[ I = \int \frac{2 - u^2}{2u^2 - 1} (-du) = \int \frac{u^2 - 2}{2u^2 - 1} du \]
To integrate this rational function, we perform algebraic manipulation (polynomial long division or splitting terms).
The numerator \( u^2 - 2 \) can be adjusted to match the denominator's leading term \( 2u^2 \):
\[ \frac{u^2 - 2}{2u^2 - 1} = \frac{1}{2} \left[ \frac{2u^2 - 4}{2u^2 - 1} \right] = \frac{1}{2} \left[ \frac{(2u^2 - 1) - 3}{2u^2 - 1} \right] \]
Simplify the fraction:
\[ I = \frac{1}{2} \int \left( 1 - \frac{3}{2u^2 - 1} \right) du = \frac{1}{2} \int 1 du - \frac{3}{2} \int \frac{1}{2u^2 - 1} du \]
Integrating the first part gives \( \frac{1}{2} u \).
For the second part, factor out 2 from the denominator:
\[ \frac{3}{2} \int \frac{1}{2(u^2 - 1/2)} du = \frac{3}{4} \int \frac{1}{u^2 - (1/\sqrt{2})^2} du \]
Using the standard integral \( \int \frac{1}{x^2 - a^2} dx = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| \), where \( a = 1/\sqrt{2} \):
\[ \text{Integral Part} = \frac{3}{4} \cdot \frac{1}{2(1/\sqrt{2})} \log \left| \frac{u - 1/\sqrt{2}}{u + 1/\sqrt{2}} \right| \]
\[ = \frac{3\sqrt{2}}{8} \log \left| \frac{\sqrt{2}u - 1}{\sqrt{2}u + 1} \right| = \frac{3}{4\sqrt{2}} \log \left| \frac{\sqrt{2}u - 1}{\sqrt{2}u + 1} \right| \]
Combining the parts:
\[ I = \frac{1}{2} u - \frac{3}{4\sqrt{2}} \log \left| \frac{\sqrt{2}u - 1}{\sqrt{2}u + 1} \right| + C \]
Substituting back \( u = \cos x \):
\[ I = \frac{1}{2} \cos x - \frac{3}{4\sqrt{2}} \log \left| \frac{\sqrt{2} \cos x - 1}{\sqrt{2} \cos x + 1} \right| + C \]
Comparing this with the given form \( P \cos x + Q \log \left| \dots \right| + C \):
We find \( P = \frac{1}{2} \) and \( Q = -\frac{3}{4\sqrt{2}} \).
Step 3: Final Answer:
Comparing the results, \( P = 1/2 \) and \( Q = -3/(4\sqrt{2}) \).
This matches Option (B).