What is the integral of (10)/(1-cos(5x)) ?

The integral of (10)/(1-cos(5x)) is -2cot((5x)/2)+C

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integral of (10)/(1-cos(5x))

\int \frac{10}{1 - cos ( 5 x )} d x

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

Solution steps

\int \frac{10}{1 - cos ( 5 x )} d x

Take the constant out:

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

= 10 \cdot \int \frac{1}{1 - cos ( 5 x )} d x

Apply u-substitution

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

= 10 \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

= 10 \cdot \frac{1}{5} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= 10 \cdot \frac{1}{5} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 10 \cdot \frac{1}{5} (- \frac{1}{v})

Substitute back

= 10 \cdot \frac{1}{5} (- \frac{1}{tan ( \frac{5 x}{2} )})

Simplify

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

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

Add a constant to the solution

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

\int \frac{( x + 1 )}{x ^{2} + 1} d x

\frac{d}{d x} (\frac{2 x}{- 8 x - 1})

y^{^{'}} - \frac{2}{t} y = t^{2} e^{t}

cos^{2} (x) sin (x) (\frac{d y}{d x}) + cos^{3} (x) y = 1

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What is the integral of (10)/(1-cos(5x)) ?
The integral of (10)/(1-cos(5x)) is -2cot((5x)/2)+C