What is the integral of (28)/(1-cos(4x)) ?

The integral of (28)/(1-cos(4x)) is -7cot(2x)+C

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integral of (28)/(1-cos(4x))

\int \frac{28}{1 - cos ( 4 x )} d x

- 7 cot (2 x) + C

Solution steps

\int \frac{28}{1 - cos ( 4 x )} d x

Take the constant out:

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

= 28 \cdot \int \frac{1}{1 - cos ( 4 x )} d x

Apply u-substitution

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

= 28 \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

= 28 \cdot \frac{1}{4} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= 28 \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 28 \cdot \frac{1}{4} (- \frac{1}{v})

Substitute back

= 28 \cdot \frac{1}{4} (- \frac{1}{tan ( \frac{4 x}{2} )})

Simplify

28 \cdot \frac{1}{4} (- \frac{1}{tan ( \frac{4 x}{2} )}) : - 7 cot (2 x)

= - 7 cot (2 x)

Add a constant to the solution

= - 7 cot (2 x) + C

y^{^{'}} = y + x, y (0) = 1

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

\int sin (t) e^{- \frac{2}{3} t} d t

\int_{1}^{2} x^{2} e^{x^{3} - 1} d x

What is the integral of (28)/(1-cos(4x)) ?
The integral of (28)/(1-cos(4x)) is -7cot(2x)+C