First slide

Methods of integration

Question

$\int \frac{\sin x}{\sin x - \cos x} d x$

Moderate

A

$\frac{1}{2} \ln |\sin x - \cos x| + \frac{x}{2} + c$
B

$\frac{1}{2} \ln |\sin x + \cos x| + c$
C

$\frac{1}{2} \ln |\sin x + 2 \cos x| + c$
D

None of these

Solution

$\begin{array}{l} Let N r = A \frac{d}{d x} (D r) + B (D r) \\ \sin x = A \frac{d}{d x} (\sin x - \cos x) + B (\sin x - \cos x) \\ \sin x = A (\cos x + \sin x) + B (\sin x - \cos x) \\ 0 . \cos x + 1 \cdot \sin x = (A - B) \cos x + (A + B) \sin x \end{array}$
$\begin{array}{l} On comparing both sides \\ A - B = 0 and A + B = 1 \\ Solving we get A = \frac{1}{2} B = \frac{1}{2} \\ \int \frac{\sin x}{\sin x - \cos x} d x \\ = \int \frac{\frac{1}{2} (\cos x + \sin x) + \frac{1}{2} (\sin x - \cos x)}{\sin x - \cos x} d x \\ = \frac{1}{2} \int \frac{\cos x + \sin x}{\sin x - \cos x} d x + \frac{1}{2} \int \frac{\sin x - \cos x}{\sin x - \cos x} d x \\ = \frac{1}{2} \log_{e} | \sin x - \cos x | + \frac{1}{2} \int 1 d x \end{array}$
$= \frac{1}{2} \ln |\sin x - \cos x| + \frac{1}{2} x + c$

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