What is the integral of (72)/(1-cos(9x)) ?

The integral of (72)/(1-cos(9x)) is -8cot((9x)/2)+C

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integral of (72)/(1-cos(9x))

\int \frac{72}{1 - cos ( 9 x )} d x

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

Solution steps

\int \frac{72}{1 - cos ( 9 x )} d x

Take the constant out:

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

= 72 \cdot \int \frac{1}{1 - cos ( 9 x )} d x

Apply u-substitution

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

= 72 \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

= 72 \cdot \frac{1}{9} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= 72 \cdot \frac{1}{9} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 72 \cdot \frac{1}{9} (- \frac{1}{v})

Substitute back

= 72 \cdot \frac{1}{9} (- \frac{1}{tan ( \frac{9 x}{2} )})

Simplify

72 \cdot \frac{1}{9} (- \frac{1}{tan ( \frac{9 x}{2} )}) : - 8 cot (\frac{9 x}{2})

= - 8 cot (\frac{9 x}{2})

Add a constant to the solution

= - 8 cot (\frac{9 x}{2}) + C

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

\frac{( 4 + x )}{( x - 2 )}, at x = 8

\frac{4}{x}

\int_{0}^{\frac{π}{2}} - 5 cos^{5} (x) d x

\frac{d}{d x} (3 - x^{2})

What is the integral of (72)/(1-cos(9x)) ?
The integral of (72)/(1-cos(9x)) is -8cot((9x)/2)+C