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

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

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integral of 6/(cos(θ)-1)

\int \frac{6}{cos ( θ ) - 1} d θ

6 cot (\frac{θ}{2}) + C

Solution steps

\int \frac{6}{cos ( θ ) - 1} d θ

Take the constant out:

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

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

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{θ}{2})

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

Simplify

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

= 6 cot (\frac{θ}{2})

Add a constant to the solution

= 6 cot (\frac{θ}{2}) + C

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