What is the integral of (16)/(1-cos(8x)) ?

The integral of (16)/(1-cos(8x)) is -2cot(4x)+C

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integral of (16)/(1-cos(8x))

\int \frac{16}{1 - cos ( 8 x )} d x

- 2 cot (4 x) + C

Solution steps

\int \frac{16}{1 - cos ( 8 x )} d x

Take the constant out:

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

= 16 \cdot \int \frac{1}{1 - cos ( 8 x )} d x

Apply u-substitution

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

= 16 \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

= 16 \cdot \frac{1}{8} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= 16 \cdot \frac{1}{8} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 16 \cdot \frac{1}{8} (- \frac{1}{v})

Substitute back

= 16 \cdot \frac{1}{8} (- \frac{1}{tan ( \frac{8 x}{2} )})

Simplify

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

= - 2 cot (4 x)

Add a constant to the solution

= - 2 cot (4 x) + C

e^{- t + 1}

\frac{d}{d x} (\frac{1}{5 x ^{5}})

\frac{- 1}{x ^{2}}

\int \frac{4}{1 + 5 x} d x

\frac{d y}{d x} = - (1 + y)^{4}, y (1) = 2

What is the integral of (16)/(1-cos(8x)) ?
The integral of (16)/(1-cos(8x)) is -2cot(4x)+C