What is the integral of 1/(cos(2z)-1) ?

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

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integral of 1/(cos(2z)-1)

integral

\frac{1}{cos ( 2 z ) - 1}

\frac{1}{2} cot (z) + C

Solution steps

\int \frac{1}{cos ( 2 z ) - 1} d z

Rewrite using trig identities

= \int - \frac{1}{2 sin ^{2} ( z )} d z

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{1}{sin ^{2} ( z )} d z

Use the common integral:

\int \frac{1}{sin ^{2} ( z )} d z = - cot (z)

= - \frac{1}{2} (- cot (z))

Simplify

= \frac{1}{2} cot (z)

Add a constant to the solution

= \frac{1}{2} cot (z) + C

\frac{v + 1}{v ^{2} + 1}

7 x^{6}

(x^{2} + a^{2})^{\frac{1}{2}}

sin (x^{2} + 1)

2 + 5 x

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