$\int \frac{{cos}^{3} x}{{sin}^{2} x + sin x} d x$ is equal to

ℓ n | sin x | + sin x + c

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ℓ n | sin x | - sin x + c

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- ℓ n | sin x | - sin x + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- ℓ n | sin x | + sin x + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is B $ℓ n | sin x | - sin x + c$
$\int \frac{{c o s}^{2} x}{{s i n}^{2} x + s i n x} d x$

$\Rightarrow \int \frac{(1 - {s i n}^{2} x) c o s x}{s i n x (s i n x + 1)} d x$

$s i n x = t$

$c o s x d x = d u$

$\Rightarrow \int \frac{(1 - t) (1 + t)}{t (t + 1)} d t$

$= \int \frac{1}{t} d t - \int 1 d t$

$= l n | t | - t + c$

$= l n | s i n x | - s i n x + c$

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