To solve the given integral equation, we need to understand and simplify the expression and find the product \( AB \). The integral is:
\[\int \frac{\sin^{\frac{2}{3}} x + \cos^{\frac{2}{3}} x}{\sqrt{\sin^{\frac{1}{3}} x \cos^{\frac{1}{3}} x \sin(x - \theta)}} \, dx = A \sqrt{\cos \theta \tan x - \sin \theta} + B \sqrt{\cos \theta \cot x + \sin(x - \theta)} + C\]Firstly, examine the expression under the integral. Simplifying or transforming parts of the integral might make it easier to evaluate or compare with the right-hand side:
Let's simplify using appropriate substitutions:
Upon simplifying, you can express the solution components separately, leading to the following steps:
Finally, calculate \( AB \) using the derived values:
\[AB = 8 \csc(2\theta)\]Therefore, the correct answer for \( AB \) is \( 8 \csc(2\theta) \).
If the roots of the quadratic equation \( ax^2 + bx + c = 0 \) are real and equal, then:
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of \( P_1 \) and \( P_2 \) are orthogonal to each other. The polarizer \( P_3 \) covers both the slits with its transmission axis at \( 45^\circ \) to those of \( P_1 \) and \( P_2 \). An unpolarized light of wavelength \( \lambda \) and intensity \( I_0 \) is incident on \( P_1 \) and \( P_2 \). The intensity at a point after \( P_3 \), where the path difference between the light waves from \( S_1 \) and \( S_2 \) is \( \frac{\lambda}{3} \), is:
