What is the integral of (63)/(1-cos(9x)) ?

The integral of (63)/(1-cos(9x)) is -7cot((9x)/2)+C

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integral of (63)/(1-cos(9x))

\int \frac{63}{1 - cos ( 9 x )} d x

- 7 cot (\frac{9 x}{2}) + C

Solution steps

\int \frac{63}{1 - cos ( 9 x )} d x

Take the constant out:

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

= 63 \cdot \int \frac{1}{1 - cos ( 9 x )} d x

Apply u-substitution

: \int \frac{1}{9 ( 1 - cos ( u ) )} d u

= 63 \cdot \int \frac{1}{9 ( 1 - cos ( u ) )} d u

Take the constant out:

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

= 63 \cdot \frac{1}{9} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= 63 \cdot \frac{1}{9} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 63 \cdot \frac{1}{9} (- \frac{1}{v})

Substitute back

= 63 \cdot \frac{1}{9} (- \frac{1}{tan ( \frac{9 x}{2} )})

Simplify

63 \cdot \frac{1}{9} (- \frac{1}{tan ( \frac{9 x}{2} )}) : - 7 cot (\frac{9 x}{2})

= - 7 cot (\frac{9 x}{2})

Add a constant to the solution

= - 7 cot (\frac{9 x}{2}) + C

\frac{\partial}{\partial u} (2 v cos (u))

f (x) = 5 x e^{x} csc (x)

\frac{d}{d x} (7 + 5 x - 5 x^{3} + \frac{7}{x ^{3}})

x \to 0 lim (- x^{2})

t \to - 5 lim (\frac{t ^{2}}{5 - t})

What is the integral of (63)/(1-cos(9x)) ?
The integral of (63)/(1-cos(9x)) is -7cot((9x)/2)+C