What is the integral of (40)/(1-cos(8x)) ?

The integral of (40)/(1-cos(8x)) is -5cot(4x)+C

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integral of (40)/(1-cos(8x))

\int \frac{40}{1 - cos ( 8 x )} d x

- 5 cot (4 x) + C

Solution steps

\int \frac{40}{1 - cos ( 8 x )} d x

Take the constant out:

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

= 40 \cdot \int \frac{1}{1 - cos ( 8 x )} d x

Apply u-substitution

: \int \frac{1}{8 ( 1 - cos ( u ) )} d u

= 40 \cdot \int \frac{1}{8 ( 1 - cos ( u ) )} d u

Take the constant out:

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

= 40 \cdot \frac{1}{8} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= 40 \cdot \frac{1}{8} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 40 \cdot \frac{1}{8} (- \frac{1}{v})

Substitute back

= 40 \cdot \frac{1}{8} (- \frac{1}{tan ( \frac{8 x}{2} )})

Simplify

40 \cdot \frac{1}{8} (- \frac{1}{tan ( \frac{8 x}{2} )}) : - 5 cot (4 x)

= - 5 cot (4 x)

Add a constant to the solution

= - 5 cot (4 x) + C

x \to \infty lim (67)

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

\frac{3}{x ^{9}}

x \to \infty lim ((x + 1)^{x + 2})

\frac{d x}{d t} = \frac{4 x ^{2} + 7 t 4 t ^{2} + x ^{2}}{4 t x}

What is the integral of (40)/(1-cos(8x)) ?
The integral of (40)/(1-cos(8x)) is -5cot(4x)+C