$\int \frac{2 s i n x}{(3 + s i n 2 x)} d x$ is equal to

$\frac{1}{2} l n ∣ ∣ \frac{2 + s i n x - c o s x}{2 - s i n x + c o s x} ∣ ∣ - \frac{1}{\sqrt{2}} t a n^{- 1} (\frac{s i n x + c o s x}{\sqrt{2}}) + C$

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$\frac{1}{2} l n ∣ ∣ \frac{2 + s i n x - c o s x}{2 - s i n x + c o s x} ∣ ∣ - \frac{1}{2 \sqrt{2}} t a n^{- 1} (\frac{s i n x + c o s x}{\sqrt{2}}) + C$

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$\frac{1}{4} l n ∣ ∣ \frac{2 + s i n x - c o s x}{2 - s i n x + c o s x} ∣ ∣ - \frac{1}{\sqrt{2}} t a n^{- 1} (\frac{s i n x + c o s x}{\sqrt{2}}) + C$

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None of these

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Solution

The correct option is C
$\frac{1}{4} l n ∣ ∣ \frac{2 + s i n x - c o s x}{2 - s i n x + c o s x} ∣ ∣ - \frac{1}{\sqrt{2}} t a n^{- 1} (\frac{s i n x + c o s x}{\sqrt{2}}) + C$

$\int \frac{2 s i n x}{(3 + s i n 2 x)} d x$
$= \int \frac{s i n x + c o s x + s i n x - c o s x}{(3 + s i n 2 x)} d x$
$= \int \frac{s i n x + c o s x}{3 + s i n 2 x} d x - \int \frac{- s i n x + c o s x}{(3 + s i n 2 x)}$
= $I_{1} + I_{2}$
Putting $t_{1} = s i n x - c o s x in I_{1} and t_{2} = s i n x + c o s x in I_{2}$ , we get
$I = \int \frac{d t_{1}}{(3 + (1 - t_{1}^{2}))} - \int \frac{d t_{2}}{(3 + (t_{2}^{2} - 1))} = \int \frac{d t_{1}}{4 - t_{1}^{2}} - \int \frac{d t_{2}}{2 + t_{2}^{2}}$
$= \frac{1}{4} l n ∣ ∣ \frac{2 + t_{1}}{2 - t_{1}} ∣ ∣ - \frac{1}{\sqrt{2}} t a n^{- 1} (\frac{t_{2}}{\sqrt{2}}) + C$
$= \frac{1}{2} l n ∣ ∣ \frac{2 + s i n x - c o s x}{2 - s i n x + c o s x} ∣ ∣ - \frac{1}{\sqrt{2}} t a n^{- 1} (\frac{s i n x + c o s x}{\sqrt{2}}) + C$

Suggest Corrections

∫2 sin x(3+sin 2x)dx is equal to

$\int \frac{2 s i n x}{(3 + s i n 2 x)} d x$ is equal to