What is the integral of 1/(3+2sin(x)) ?

The integral of 1/(3+2sin(x)) is 2/(sqrt(5))arctan(1/(sqrt(5))(3tan(x/2)+2))+C

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integral of 1/(3+2sin(x))

\int \frac{1}{3 + 2 sin ( x )} d x

\frac{2}{5} arctan (\frac{1}{5} (3 tan (\frac{x}{2}) + 2)) + C

Solution steps

\int \frac{1}{3 + 2 sin ( x )} d x

Apply u-substitution

: \int \frac{2}{3 ( 1 + u ^{2} ) + 4 u} d u

= \int \frac{2}{3 ( 1 + u ^{2} ) + 4 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{3 ( 1 + u ^{2} ) + 4 u} d u

Expand

3 (1 + u^{2}) : 3 + 3 u^{2}

= 2 \cdot \int \frac{1}{3 + 3 u ^{2} + 4 u} d u

Complete the square

3 + 3 u^{2} + 4 u : 3 (u + \frac{2}{3})^{2} + \frac{5}{3}

= 2 \cdot \int \frac{1}{3 ( u + \frac{2}{3} ) ^{2} + \frac{5}{3}} d u

Apply u-substitution

: \int \frac{3}{9 v ^{2} + 5} d v

= 2 \cdot \int \frac{3}{9 v ^{2} + 5} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot 3 \cdot \int \frac{1}{9 v ^{2} + 5} d v

Apply Integral Substitution

= 2 \cdot 3 \cdot \int \frac{1}{3 5 ( w ^{2} + 1 )} d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot 3 \cdot \frac{1}{3 5} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= 2 \cdot 3 \cdot \frac{1}{3 5} arctan (w)

Substitute back

= 2 \cdot 3 \cdot \frac{1}{3 5} arctan (\frac{3}{5} (tan (\frac{x}{2}) + \frac{2}{3}))

Simplify

2 \cdot 3 \cdot \frac{1}{3 5} arctan (\frac{3}{5} (tan (\frac{x}{2}) + \frac{2}{3})) : \frac{2}{5} arctan (\frac{1}{5} (3 tan (\frac{x}{2}) + 2))

= \frac{2}{5} arctan (\frac{1}{5} (3 tan (\frac{x}{2}) + 2))

Add a constant to the solution

= \frac{2}{5} arctan (\frac{1}{5} (3 tan (\frac{x}{2}) + 2)) + C

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What is the integral of 1/(3+2sin(x)) ?
The integral of 1/(3+2sin(x)) is 2/(sqrt(5))arctan(1/(sqrt(5))(3tan(x/2)+2))+C