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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Complete the square

3 (1 + u^{2}) + 2 u : 3 (u + \frac{1}{3})^{2} + \frac{8}{3}

= 2 \cdot \int \frac{1}{3 ( u + \frac{1}{3} ) ^{2} + \frac{8}{3}} d u

Apply u-substitution

= 2 \cdot \int \frac{3}{9 v ^{2} + 8} 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} + 8} d v

Apply Integral Substitution

= 2 \cdot 3 \cdot \int \frac{1}{6 2 ( 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}{6 2} \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}{6 2} arctan (w)

Substitute back

= 2 \cdot 3 \cdot \frac{1}{6 2} arctan (\frac{3}{2 2} (tan (\frac{x}{2}) + \frac{1}{3}))

Simplify

2 \cdot 3 \cdot \frac{1}{6 2} arctan (\frac{3}{2 2} (tan (\frac{x}{2}) + \frac{1}{3})) : \frac{1}{2} arctan (\frac{1}{2 2} (3 tan (\frac{x}{2}) + 1))

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

Add a constant to the solution

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

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