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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

= \frac{1}{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

= \frac{1}{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}

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{w}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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