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

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

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integral of 1/(2-sin(x))

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 1 - u} d u

= \int \frac{1}{u ^{2} + 1 - u} d u

Complete the square

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

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

Apply u-substitution

: \int \frac{4}{4 v ^{2} + 3} d v

= \int \frac{4}{4 v ^{2} + 3} d v

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{1}{1 + 2 + n}

\int \frac{( x - 3 )}{x + 3} d x

x^{2} - 11, - 2

\frac{d}{d x} (\frac{x}{4})

\int x^{- \frac{4}{3}} d x

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