What is the integral of 8/(1-cos(2x)) ?

The integral of 8/(1-cos(2x)) is -4cot(x)+C

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integral of 8/(1-cos(2x))

\int \frac{8}{1 - cos ( 2 x )} d x

- 4 cot (x) + C

Solution steps

\int \frac{8}{1 - cos ( 2 x )} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{1 - cos ( 2 x )} d x

Rewrite using trig identities

= 8 \cdot \int \frac{1}{2 sin ^{2} ( x )} d x

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{sin ^{2} ( x )} d x

Use the common integral:

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

= 8 \cdot \frac{1}{2} (- cot (x))

Simplify

8 \cdot \frac{1}{2} (- cot (x)) : - 4 cot (x)

= - 4 cot (x)

Add a constant to the solution

= - 4 cot (x) + C

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What is the integral of 8/(1-cos(2x)) ?
The integral of 8/(1-cos(2x)) is -4cot(x)+C