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
⇒ ∫ [sin2 x + cos2 x + 2cos x] / (2 + cos x)2 dx
⇒ ∫ [cos x (cos x + 2) + sin2 x] / (2 + cos x)2 dx
⇒ ∫ cos x dx / (cos x + 2) + ∫ sin2 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] + ∫ sin2 x dx / (2 + cos x)2
⇒ sin x / (cos x + 2) – ∫[−(− sin x) sin x dx] / (2 + cos x)2 + ∫ sin2 x dx / (2 + cos x)2
⇒ sin x / (cos x + 2) – ∫ sin2 x dx / (2 + cos x)2 + ∫ sin2 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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