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 : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,