What is the integral of (63)/(1-cos(7x)) ?

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

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integral of (63)/(1-cos(7x))

\int \frac{63}{1 - cos ( 7 x )} d x

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

Solution steps

\int \frac{63}{1 - cos ( 7 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 ( 7 x )} d x

Apply u-substitution

: \int \frac{1}{7 ( 1 - cos ( u ) )} d u

= 63 \cdot \int \frac{1}{7 ( 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}{7} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

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

= 63 \cdot \frac{1}{7} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 63 \cdot \frac{1}{7} (- \frac{1}{v})

Substitute back

= 63 \cdot \frac{1}{7} (- \frac{1}{tan ( \frac{7 x}{2} )})

Simplify

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

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

Add a constant to the solution

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

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