First slide

Evaluation of definite integrals

Question

The value of $\int_{0}^{π / 2} \frac{1 + 2 \cos x}{(2 + \cos x)^{2}} d x$ is

Difficult

A

-1/2
B

2
C

1/2
D

none of these

Solution

$\int_{0}^{π / 2} \frac{1 + 2 \cos x}{(2 + \cos x)^{2}} d x = \int_{0}^{π / 2} \frac{2 (\cos x + 2) - 3}{(2 + \cos x)^{2}} d x$
$\begin{array}{l} = 4 \int_{0}^{1} \frac{d t}{3 + t^{2}} - 6 \int_{0}^{1} \frac{1 + t^{2}}{{(3 + t^{2})}^{2}} d t (t = \tan x / 2) \\ = - 2 \int_{0}^{1} \frac{d t}{3 + t^{2}} + 12 \int_{0}^{1} \frac{d t}{{(3 + t^{2})}^{2}} \\ = - 2 \int_{0}^{1} \frac{d t}{3 + t^{2}} + 12 [{\frac{1}{2.3} \frac{t}{(t^{2} + 3)}|}_{0}^{1} + \frac{1}{2.3} \int_{0}^{1} \frac{d t}{3 + t^{2}}] \\ = \frac{1}{2} . \end{array}$

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