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Introduction to integration

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Question

$Let g (x) = \int \frac{1 + 2 \cos x}{(\cos x + 2)^{2}} d x and g (0) = 0 . Then the value of g (3 π / 2) is$

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Solution

$\begin{array}{l} g (x) = \int \frac{1 + 2 \cos x}{(\cos x + 2)^{2}} d x \\ = \int \frac{{cosec}^{2} x + 2 \cot x cosec x}{(\cot x + 2 cosec x)^{2}} d x \\ (Dividing numerator and denominator by \sin^{2} x) \end{array}$
$\begin{array}{l} = \frac{1}{\cot x + 2 cosec x} + C \\ g (x) = \frac{\sin x}{\cos x + 2} + C \\ Since g (0) = 0, we get that C = 0 \\ So, g (x) = \frac{\sin x}{\cos x + 2} \\ ∴ g (3 π / 2) = - \frac{1}{2} \end{array}$

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