-d x-sin x-sin 2 x-A- 1-6log 1-cos x-1-2log 1-cos x-2-3log 1-2 cos x- cB- 1-6log 1-sin x-1-2log 1-cos x-2-3log 1-2 sin x- cC- 1-6log 1-sin x-1-2log 1-sin x-2-3log 1-2 sin x- cD- 1-2log 1-sin x-2-3log 1-cos x-1-6log 1-2 cos x- c

The correct option is A 1/6log(1 cos x)+1/2log(1+cos x) 2/3log(1+2cos x)+c∫dx/sin x (1+2 cos x) =∫sin⁡ x dx/sin^2x(1+2cos x)=∫sin⁡ x dx/(1 cos x)(1+cos x)(1+2c ...

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Byju's Answer

Standard XII

Mathematics

Area between x=g(y) and y Axis

∫d x/sin x+si...

Question

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

\frac{1}{6} l o g (1 - c o s x) + \frac{1}{2} l o g (1 + c o s x) - \frac{2}{3} l o g (1 + 2 c o s x) + c

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\frac{1}{6} l o g (1 - s i n x) + \frac{1}{2} l o g (1 + c o s x) - \frac{2}{3} l o g (1 + 2 s i n x) + c

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\frac{1}{6} l o g (1 - s i n x) + \frac{1}{2} l o g (1 + s i n x) - \frac{2}{3} l o g (1 + 2 s i n x) + c

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\frac{1}{2} l o g (1 - s i n x) + \frac{2}{3} l o g (1 + c o s x) - \frac{1}{6} l o g (1 + 2 c o s x) + c

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Solution

The correct option is A $\frac{1}{6} l o g (1 - c o s x) + \frac{1}{2} l o g (1 + c o s x) - \frac{2}{3} l o g (1 + 2 c o s x) + c$
$\int \frac{d x}{s i n x (1 + 2 c o s x)}$
$= \int \frac{s i n x d x}{s i n^{2} x (1 + 2 c o s x)} = \int \frac{s i n x d x}{(1 - c o s x) (1 + c o s x) (1 + 2 c o s x)}$
Let cos x = t
$∴ \int \frac{- d t}{(1 - t) (1 + t) (1 + 2 t)} = \int (\frac{- 1}{6 (1 - t)} + \frac{1}{2 (1 + t)} - \frac{4}{3 (1 + 2 t)}) d t$
$= \frac{1}{6} l o g (1 - t) + \frac{1}{2} l o g (1 + t) - \frac{2}{3} l o g (1 + 2 t)$ with $t = c o s x$

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