We are tasked with evaluating the integral: \[ I = \int_0^{\frac{\pi}{2}} \sin^2(x) \, dx. \]
We use the identity for \( \sin^2(x) \): \[ \sin^2(x) = \frac{1 - \cos(2x)}{2}. \]
Thus, the integral becomes: \[ I = \int_0^{\frac{\pi}{2}} \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \int_0^{\frac{\pi}{2}} (1 - \cos(2x)) \, dx. \]
We can now split the integral: \[ I = \frac{1}{2} \left[ \int_0^{\frac{\pi}{2}} 1 \, dx - \int_0^{\frac{\pi}{2}} \cos(2x) \, dx \right]. \]
Evaluating each integral: \[ \int_0^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2}, \quad \int_0^{\frac{\pi}{2}} \cos(2x) \, dx = 0. \]
Therefore, the value of the integral is: \[ I = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}. \]
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32