First slide

Methods of integration

Question

$\int \frac{\sin 2 x}{\sin 5 x \sin 3 x} d x =$

Moderate

A

$\log |\sin x| - \log |\cos x| + c$
B

$\frac{1}{2} \log |\sin 2 x| + \frac{1}{3} \log |\sin 3 x| + c$
C

$\frac{1}{3} \log |\sin 2 x| + \frac{1}{5} \log |\sin 3 x| + c$
D

$\frac{1}{3} \log |\sin 3 x| - \frac{1}{5} \log |\sin 5 x| + c$

Solution

$\int \frac{\sin 2 x}{\sin 5 x \sin 3 x} d x$
$\begin{array}{l} = \int \frac{\sin (5 x - 3 x)}{\sin 5 x \cdot \sin 3 x} d x \\ = \int \frac{\sin 5 x \cos 3 x - \cos 5 x \sin 3 x}{\sin 5 x \cdot \sin 3 x} d x \\ = \int \cot 3 x d x - \int \cot 5 x d x \\ = \frac{1}{3} \log | \sin 3 x | - \frac{1}{5} | \sin 5 x | + c \end{array}$

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