First slide

Methods of integration

Question

$\int \frac{d x}{5 + 4 \cos x} = a \tan^{- 1} (b \tan \frac{x}{2}) + C, then$

Moderate

A

$a = \frac{2}{3}, b = - \frac{1}{3}$
B

$a = \frac{2}{3}, b = \frac{1}{3}$
C

$a = - \frac{2}{3}, b = \frac{1}{3}$
D

$a = - \frac{2}{3}, b = - \frac{1}{3}$

Solution

$\begin{array}{l} I = \int \frac{d x}{5 + 4 \cos x} \\ = \int \frac{d x}{5 (\sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2}) + 4 (\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2})} \\ = \int \frac{d x}{9 \cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2}} = \int \frac{\sec^{2} \frac{x}{2}}{9 + \tan^{2} \frac{x}{2}} d x \end{array}$
$\begin{array}{l} Now, I = \int \frac{2 d t}{9 + t^{2}} = \frac{2}{3} \tan^{- 1} (\frac{t}{3}) + C \\ ∴ I = \frac{2}{3} \tan^{- 1} (\frac{\tan (\frac{x}{2})}{3}) + C \end{array}$
$Hence, a = \frac{2}{3}, b = \frac{1}{3}$

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