$\int \frac{cos x}{(1 + sin x) (2 + sin x)} d x =$

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

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

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log | (1 + s i n x) (2 + sin x) | + c

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log | (1 + s i n x) + (2 + sin x) | + c

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Solution

The correct option is A $log | \frac{1 + sin x}{2 + sin x} | + c$
$\int \frac{d s i n x}{(2 + s i n x) (1 + s i n x)}$
$= \int \frac{1}{(s i n x + 1)} d s i n x - \int \frac{d s i n x}{(s i n x + 2)}$
$= l o g ∣ s i n x + 1 ∣ - l o g ∣ s i n x + 2 ∣ + c$
$= l o g ∣ \frac{s i n x + 1}{s i n x + 2} ∣ + c$

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Similar questions

\int \frac{sin 2 x}{(1 + sin x) (2 + sin x)} d x = - 2 log | 1 + sin x | + m log | 2 + sin x | + c

m

\int \frac{sin x}{sin (x - π / 4)} d x = A (f (x)) + log | sin x - cos x | + C

If ${log}_{| s i n x |} | c o s x | + {log}_{| c o s x |} | s i n x | = 2$ then |tanx|

[Hint: Put sin x = t]

y = log (\sqrt{\frac{1 + sin x}{1 - sin x}})

\frac{d y}{d x}

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