Evaluate the integral: int_0^{pi /2} frac{1 + | vedclass

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(C) Let $I = \int_0^{\pi /2} \frac{1 + 2\cos x}{(2 + \cos x)^2} dx$. We can rewrite the numerator as $2(2 + \cos x) - 3$. So,$I = \int_0^{\pi /2} \fra

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) Let $I = \int_0^{\pi /2} \frac{1 + 2\cos x}{(2 + \cos x)^2} dx$. We can rewrite the numerator as $2(2 + \cos x) - 3$. So,$I = \int_0^{\pi /2} \frac{2(2 + \cos x) - 3}{(2 + \cos x)^2} dx = 2 \int_0^{\pi /2} \frac{1}{2 + \cos x} dx - 3 \int_0^{\pi /2} \frac{1}{(2 + \cos x)^2} dx$. Using the substitution $t = 	an(x/2)$,we have $\cos x = \frac{1 - t^2}{1 + t^2}$ and $dx = \frac{2 dt}{1 + t^2}$. As $x$ goes from $0$ to $\pi/2$,$t$ goes from $0$ to $1$. $I = 2 \int_0^1 \frac{1}{2 + \frac{1 - t^2}{1 + t^2}} \cdot \frac{2 dt}{1 + t^2} - 3 \int_0^1 \frac{1}{(2 + \frac{1 - t^2}{1 + t^2})^2} \cdot \frac{2 dt}{1 + t^2} = 4 \int_0^1 \frac{dt}{3 + t^2} - 6 \int_0^1 \frac{1 + t^2}{(3 + t^2)^2} dt$. Using integration by parts on the second integral or reduction formulas,we find the result is $\frac{1}{2}$.

Evaluate the integral: $\int_0^{\pi /2} \frac{1 + 2\cos x}{(2 + \cos x)^2} dx$

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