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 \( 8 \csc(2\theta) \).
Let p and q be two real numbers such that p + q = 3 and p4 + q4 = 369. Then
\((\frac{1}{p} + \frac{1}{q} )^{-2}\)
is equal to _______.
Let \( \alpha, \beta; \, \alpha > \beta \), be the roots of the equation $$ x^2 - \sqrt{2}x - \sqrt{3} = 0. $$ Let \( P_n = \alpha^n - \beta^n, \, n \in \mathbb{N} \). Then $$ \left( 11\sqrt{3} - 10\sqrt{2} \right) P_{10} + \left( 11\sqrt{2} + 10 \right) P_{11} - 11P_{12} $$ is equal to:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)