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

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

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

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

- \frac{1}{2} cot (x) + C

Solution steps

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

Rewrite using trig identities

= \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

= \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)

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

Simplify

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

Add a constant to the solution

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

\int_{0}^{4} 4 - x^{2} d x

\frac{d x}{d t} - 3 x = t

\frac{d}{d x} (\frac{4 x}{1 + 2 x ^{2}})

f (x) = x^{2} + 4, (- 5, 29)

\frac{d}{d x} (x 2 x - 3)

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