What is the integral of (32)/(1-cos(4x)) ?

The integral of (32)/(1-cos(4x)) is -8cot(2x)+C

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integral of (32)/(1-cos(4x))

\int \frac{32}{1 - cos ( 4 x )} d x

- 8 cot (2 x) + C

Solution steps

\int \frac{32}{1 - cos ( 4 x )} d x

Take the constant out:

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

= 32 \cdot \int \frac{1}{1 - cos ( 4 x )} d x

Apply u-substitution

: \int \frac{1}{4 ( 1 - cos ( u ) )} d u

= 32 \cdot \int \frac{1}{4 ( 1 - cos ( u ) )} d u

Take the constant out:

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

= 32 \cdot \frac{1}{4} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= 32 \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 32 \cdot \frac{1}{4} (- \frac{1}{v})

Substitute back

= 32 \cdot \frac{1}{4} (- \frac{1}{tan ( \frac{4 x}{2} )})

Simplify

32 \cdot \frac{1}{4} (- \frac{1}{tan ( \frac{4 x}{2} )}) : - 8 cot (2 x)

= - 8 cot (2 x)

Add a constant to the solution

= - 8 cot (2 x) + C

\int_{0}^{4} \frac{x}{x ^{2} - 4} d x

\frac{\partial}{\partial x} (r (θ, x) cos (θ))

x \to 3 - lim (\frac{3}{x ^{2} - 9})

\int_{1}^{3} \frac{x ^{2} + 5}{4 x - x ^{2}} d x

\frac{d}{d x} ((\frac{1}{1 - x})^{2})

What is the integral of (32)/(1-cos(4x)) ?
The integral of (32)/(1-cos(4x)) is -8cot(2x)+C