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Find the value of the integral (1 + 2cos x) / (2 + cos x)^{2 }dx from 0 to π/2

Integration is the inverse of differentiation. Given a derivative of a function, the process of finding the original function is called integration.

**Answer:** The value of the integral (1 + 2cos x)/(2 + cos x)^{2}^{ }dx from 0 to π/2 is 1 / 2.

Let's look into the solution step by step.

**Explanation:**

Consider, I = ∫ (1 + 2cos x) / (2 + cos x)^{2}

⇒ ∫ [sin^{2 }x + cos^{2 }x + 2cos x] / (2 + cos x)^{2 }dx

⇒ ∫ [cos x (cos x + 2) + sin^{2 }x] / (2 + cos x)^{2 }dx

⇒ ∫ cos x dx / (cos x + 2) + ∫ sin^{2 }x dx / (2 + cos x)^{2}

Now, we will integrate the first integral using uv rule

u = 1/(cos x + 2) and v = cos x

⇒ [1 / (cos x + 2)] ∫ cos x dx – ∫[1 / (cos x + 2) ∫ cos x dx] + ∫ sin^{2 }x dx / (2 + cos x)^{2}

⇒ sin x / (cos x + 2) – ∫[−(− sin x) sin x dx] / (2 + cos x)^{2} + ∫ sin^{2 }x dx / (2 + cos x)^{2}

⇒ sin x / (cos x + 2) – ∫ sin^{2 }x dx / (2 + cos x)^{2} + ∫ sin^{2 }x dx / (2 + cos x)^{2}

⇒ sin x / (cos x + 2) + c

Now let's apply the limits,

\(\int\limits_0^{\pi/2}\) (1 + 2cos x) / (2 + cos x)^{2}

⇒ sin x / (cos x +2) \(\left.\right|_0^{\pi/2}\)

⇒ [sin π/2 / (cos π/2 + 2)] - [sin 0 / (cos 0 + 2)]

⇒ [1 / (0 + 2)] - [0 / (1 + 2)]

⇒ (1 / 2) - 0

⇒ 1 / 2

### Thus, the value of the integral (1 + 2cos x) / (2 + cos x)^{2 }dx from 0 to π/2 is 1/2.

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