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

\int 24 arctan (x) d x

\int sec^{6} (2 x) tan (2 x) d x

cos (x) \frac{d y}{d x} - sin (x) y = 0

\frac{\partial}{\partial x} (\frac{x - y}{x y + 2})

x \to \infty lim (ln (1))

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