First slide

Methods of integration

Question

$If \int \frac{3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x} d x = a x + b \ln |2 \sin x + 3 \cos x| + C then$

Moderate

A

$a = - \frac{12}{13}, b = \frac{15}{39}$
B

$a = - \frac{7}{13}, b = \frac{6}{13}$
C

$a = \frac{12}{13}, b = - \frac{15}{39}$
D

$a = - \frac{7}{13}, b = - \frac{6}{13}$

Solution

Differentiating both sides, we get
$\begin{aligned} \frac{3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x} = a + \frac{b (2 \cos x - 3 \sin x)}{(2 \sin x + 3 \cos x)} \\ = \frac{\sin x \cdot (2 a - 3 b) + \cos x \cdot (3 a + 2 b)}{(3 \cos x + 2 \sin x)} \end{aligned}$
Comparing like terms on both sides, we get
$3 = 2 a - 3 b, 2 = 3 a + 2 b \Rightarrow a = \frac{12}{13}, b = - \frac{15}{39}$

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