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

The integral of 1/(1-sin(x/2)) is -4/(tan(x/4)-1)+C

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

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

- \frac{4}{tan ( \frac{x}{4} ) - 1} + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Complete the square

1 + v^{2} - 2 v : (v - 1)^{2}

= 2 \cdot 2 \cdot \int \frac{1}{( v - 1 ) ^{2}} d v

Apply u-substitution

: \int \frac{1}{w ^{2}} d w

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

Apply the Power Rule

: - \frac{1}{w}

= 2 \cdot 2 (- \frac{1}{w})

Substitute back

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

Simplify

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

= - \frac{4}{tan ( \frac{x}{4} ) - 1}

Add a constant to the solution

= - \frac{4}{tan ( \frac{x}{4} ) - 1} + C

f (x) = \frac{10}{x ^{2}}

\int \frac{1}{x 4 ^{2} - 9} d x

\int \frac{1}{1 - \frac{1}{2} x} d x

\frac{x}{6}

(x tan (\frac{y}{x}) + y) \cdot d x - x \cdot d y = 0

What is the integral of 1/(1-sin(x/2)) ?
The integral of 1/(1-sin(x/2)) is -4/(tan(x/4)-1)+C