We need to evaluate $ \displaystyle \int_{0}^{\pi/2} \frac{1 + 2\cos x}{(2 + \cos x)^2}\, dx $.

Let $ t = 2 + \cos x $. Then $ dt = -\sin x\, dx $.

But the integral has no $\sin x$, so rewrite numerator:

$ 1 + 2\cos x = (2 + \cos x) - 1 = t - 1 $.

Now express $ dx $ using $ \sin^2 x = 1 - \cos^2 x $, but the standard trick is to differentiate:

$ \dfrac{d}{dx}\left(\dfrac{1}{2+\cos x}\right) = -\dfrac{-\sin x}{(2+\cos x)^2} = \dfrac{\sin x}{(2+\cos x)^2}. $ We

use complementary substitution:

Let $ x = \frac{\pi}{2} - y $.

Then $\cos x = \sin y$ and $\sin x = \cos y$.

Integral becomes:

$ I = \int_{0}^{\pi/2} \frac{1 + 2\sin y}{(2 + \sin y)^2}\, dy. $

Average the two forms:

$ I = \frac{1}{2}\int_{0}^{\pi/2} \left[ \frac{1 + 2\cos x}{(2 + \cos x)^2} + \frac{1 + 2\sin x}{(2 + \sin x)^2} \right] dx. $

Now observe identity:

$ \frac{1 + 2\cos x}{(2 + \cos x)^2} + \frac{1 + 2\sin x}{(2 + \sin x)^2} = \frac{d}{dx}\left(\frac{\sin x - \cos x}{(2+\cos x)(2+\sin x)}\right). $

Thus integral becomes a telescoping form and evaluates to:

$ I = \left[ \frac{\sin x - \cos x}{(2+\cos x)(2+\sin x)} \right]_{0}^{\pi/2}. $

Now compute:

At $ x = \frac{\pi}{2}$: $ \sin x = 1,\;\cos x = 0 $

Expression = $ \dfrac{1 - 0}{(2+0)(2+1)} = \dfrac{1}{6}. $

At $ x = 0$: $ \sin 0 = 0,\;\cos 0 = 1 $

Expression = $ \dfrac{0 - 1}{(2+1)(2+0)} = -\dfrac{1}{6}. $

Therefore: $ I = \dfrac{1}{6} - (-\dfrac{1}{6}) = \dfrac{2}{6} = \dfrac{1}{3}. $

Final Answer: $\dfrac{1}{3} $

Question Practice

Discussion