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

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

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

integral

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply u-substitution

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

Complete the square

v^{2} + 1 + v : (v + \frac{1}{2})^{2} + \frac{3}{4}

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

Apply u-substitution

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{3} \cdot 4 \cdot \frac{1}{2 3} arctan (t)

Substitute back

= \frac{1}{3} \cdot 4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (tan (\frac{3 x}{2}) + \frac{1}{2}))

Simplify

\frac{1}{3} \cdot 4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (tan (\frac{3 x}{2}) + \frac{1}{2})) : \frac{2}{3 3} arctan (\frac{1}{3} (2 tan (\frac{3 x}{2}) + 1))

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

Add a constant to the solution

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

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