What is the integral of 6/(1-cos(2x)) ?

The integral of 6/(1-cos(2x)) is -3cot(x)+C

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integral of 6/(1-cos(2x))

\int \frac{6}{1 - cos ( 2 x )} d x

- 3 cot (x) + C

Solution steps

\int \frac{6}{1 - cos ( 2 x )} 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}{1 - cos ( 2 x )} d x

Rewrite using trig identities

= 6 \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

= 6 \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)

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

Simplify

6 \cdot \frac{1}{2} (- cot (x)) : - 3 cot (x)

= - 3 cot (x)

Add a constant to the solution

= - 3 cot (x) + C

x y^{^{'}} = y

x y^{^{'}} e^{x} + 2 y e^{x} = 0

\frac{\partial}{\partial x} (ln (x^{3} + y))

- 3 x^{2}, at x = 2

\int (3 x + 5)^{2} d x

What is the integral of 6/(1-cos(2x)) ?
The integral of 6/(1-cos(2x)) is -3cot(x)+C