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

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

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

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

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

Solution steps

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

Simplify

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

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\int 3 x^{3} cos (5 x) d x

x \to 5 lim (x^{2} + 11)

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