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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Complete the square

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

= 2 (- \frac{1}{v})

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

x \to 0 lim (\frac{sin ( 2 ) ( x )}{x})

\frac{1}{u ^{2}}

\frac{( 6 + 7 x )}{( x ^{4} + 5 x - 5 )}

k = 2 \sum \infty \frac{8}{4 ^{2 k}}

f (x) = - x + 3 sin (x)

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