What is the integral of (20)/(1-cos(4x)) ?

The integral of (20)/(1-cos(4x)) is -5cot(2x)+C

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integral of (20)/(1-cos(4x))

\int \frac{20}{1 - cos ( 4 x )} d x

- 5 cot (2 x) + C

Solution steps

\int \frac{20}{1 - cos ( 4 x )} d x

Take the constant out:

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

= 20 \cdot \int \frac{1}{1 - cos ( 4 x )} d x

Apply u-substitution

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

= 20 \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

= 20 \cdot \frac{1}{4} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= 20 \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 20 \cdot \frac{1}{4} (- \frac{1}{v})

Substitute back

= 20 \cdot \frac{1}{4} (- \frac{1}{tan ( \frac{4 x}{2} )})

Simplify

20 \cdot \frac{1}{4} (- \frac{1}{tan ( \frac{4 x}{2} )}) : - 5 cot (2 x)

= - 5 cot (2 x)

Add a constant to the solution

= - 5 cot (2 x) + C

x \to 4 lim (\frac{x + 5}{x - 4})

x \to \frac{9}{2} lim (2 x - 7)

\frac{\partial}{\partial y} (y^{2} sin (2 x))

1 - 2^{x}

y = x^{2} (2 x + 3)^{5}

What is the integral of (20)/(1-cos(4x)) ?
The integral of (20)/(1-cos(4x)) is -5cot(2x)+C