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

\frac{1}{2} log | sin

x-cos

x | - \frac{x}{2} + c

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\frac{1}{2} log | sin

x-cos

x | + \frac{x}{2} + c

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log | sin

x-cos

x | + x + c

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log | sin

x-cos

x | - x + c

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Solution

The correct option is A $\frac{1}{2} log | sin$ x-cos $x | - \frac{x}{2} + c$
$N (x) = c o s x; D (x) = s i n x - c o s x$
$N (x) = λ D^{1} (x) + μ (D (x))$
$c o s x = λ (c o s x + s i n x) + μ (s i n x - c o s x)$
$\begin{matrix} λ - μ = 1 λ + μ = 0 \end{matrix}} λ = \frac{1}{2}; μ = \frac{- 1}{2}$
$N (x) = \frac{1}{2} D^{1} (x) - \frac{1}{2} D (x)$
$\int \frac{N (x)}{D (x)} d x = \frac{1}{2} \int \frac{D^{1} (x)}{D (x)} - \frac{1}{2} \int \frac{D (x)}{D (x)} d x$
$= \frac{1}{2} log | D (x) | - \frac{1}{2} x + c$
$= \frac{1}{2} log | (s i n x - c o s x) | - \frac{1}{2} x + c$
$\int \frac{c o s x}{s i n x - c o s x} d x = \frac{1}{2} l o g | s i n x - c o s x | - \frac{1}{2} x + c$

Suggest Corrections

Similar questions

\int cos (log x) d x = F (x) + c,

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F (x) =

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\sqrt{\sin 2 x} + C

\sqrt{\cos 2 x} + C

y = x^{sin x}

\int \frac{d x}{s i n x + s i n 2 x} =

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