The value of the int_{0}^{frac{pi}{2}} left( | vedclass

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(D) We know that $\sin 3y = 3\sin y - 4\sin^3 y$. Substituting this into the integrand: $\frac{1 + 3\sin y - 4\sin^3 y}{1 + 2\sin y}$. Using polynomia

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(D) We know that $\sin 3y = 3\sin y - 4\sin^3 y$. Substituting this into the integrand: $\frac{1 + 3\sin y - 4\sin^3 y}{1 + 2\sin y}$. Using polynomial division or factoring: $1 + 3\sin y - 4\sin^3 y = (1 + 2\sin y)(1 + \sin y - 2\sin^2 y)$. Thus,the integrand simplifies to $1 + \sin y - 2\sin^2 y$. Now,integrate with respect to $y$ from $0$ to $\frac{\pi}{2}$: $\int_{0}^{\frac{\pi}{2}} (1 + \sin y - 2\sin^2 y) dy = \int_{0}^{\frac{\pi}{2}} (1 + \sin y - (1 - \cos 2y)) dy$. $= \int_{0}^{\frac{\pi}{2}} (\sin y + \cos 2y) dy$. $= [-\cos y + \frac{1}{2}\sin 2y]_{0}^{\frac{\pi}{2}}$. $= (-\cos \frac{\pi}{2} + \frac{1}{2}\sin \pi) - (-\cos 0 + \frac{1}{2}\sin 0)$. $= (0 + 0) - (-1 + 0) = 1$.

The value of the $\int_{0}^{\frac{\pi}{2}} \left( \frac{1 + \sin 3y}{1 + 2\sin y} \right) dy$ is equal to

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