What is the integral of 6/(cos(x)-1) ?

The integral of 6/(cos(x)-1) is 6cot(x/2)+C

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integral of 6/(cos(x)-1)

\int \frac{6}{cos ( x ) - 1} d x

6 cot (\frac{x}{2}) + C

Solution steps

\int \frac{6}{cos ( x ) - 1} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{cos ( x ) - 1} d x

Apply u-substitution

: \int - \frac{1}{u ^{2}} d u

= 6 \cdot \int - \frac{1}{u ^{2}} d u

Take the constant out:

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

= 6 (- \int \frac{1}{u ^{2}} d u)

Apply the Power Rule

: - \frac{1}{u}

= 6 (- (- \frac{1}{u}))

Substitute back

u = tan (\frac{x}{2})

= 6 (- (- \frac{1}{tan ( \frac{x}{2} )}))

Simplify

6 (- (- \frac{1}{tan ( \frac{x}{2} )})) : 6 cot (\frac{x}{2})

= 6 cot (\frac{x}{2})

Add a constant to the solution

= 6 cot (\frac{x}{2}) + C

2 y^{^{''}} + 3 y^{^{'}} - 2 y = cos (2 x)

\frac{x}{x ^{2}}

\frac{\partial}{\partial x} (\frac{9}{x} + \frac{3}{y} + x y)

\frac{d}{d x} (x^{a r c t a n (x)})

\frac{d y}{d x} + 6 (tan (6 x)) y = 9 cos (6 x)

What is the integral of 6/(cos(x)-1) ?
The integral of 6/(cos(x)-1) is 6cot(x/2)+C