What is the integral of (20)/(1-cos(5x)) ?

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

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integral of (20)/(1-cos(5x))

\int \frac{20}{1 - cos ( 5 x )} d x

- 4 cot (\frac{5 x}{2}) + C

Solution steps

\int \frac{20}{1 - cos ( 5 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 ( 5 x )} d x

Apply u-substitution

: \int \frac{1}{5 ( 1 - cos ( u ) )} d u

= 20 \cdot \int \frac{1}{5 ( 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}{5} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= 20 \cdot \frac{1}{5} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 20 \cdot \frac{1}{5} (- \frac{1}{v})

Substitute back

= 20 \cdot \frac{1}{5} (- \frac{1}{tan ( \frac{5 x}{2} )})

Simplify

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

= - 4 cot (\frac{5 x}{2})

Add a constant to the solution

= - 4 cot (\frac{5 x}{2}) + C

x \to \frac{π}{6} lim (cos (x))

t \to 1 lim (t + 2)

\frac{d}{d x} (\frac{1}{cot ( x )})

\frac{d}{d x} ((- 4 x^{2} + 9)^{2})

\frac{d}{d y} (x cos (x) y)

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