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

\frac{\partial}{\partial x} (9 - x^{2} - y^{2})

\frac{d}{d x} (ln (\frac{x ^{2}}{2 x ^{2} + 3 x + 2}))

y = \frac{π}{( 2 x ) ^{4}}

\int \frac{2}{x ^{2} + 9} d x

\frac{x d y}{d x} = 2 y

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