What is the integral of (18)/(1-cos(2x)) ?

The integral of (18)/(1-cos(2x)) is -9cot(x)+C

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integral of (18)/(1-cos(2x))

\int \frac{18}{1 - cos ( 2 x )} d x

- 9 cot (x) + C

Solution steps

\int \frac{18}{1 - cos ( 2 x )} d x

Take the constant out:

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

= 18 \cdot \int \frac{1}{1 - cos ( 2 x )} d x

Rewrite using trig identities

= 18 \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

= 18 \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)

= 18 \cdot \frac{1}{2} (- cot (x))

Simplify

18 \cdot \frac{1}{2} (- cot (x)) : - 9 cot (x)

= - 9 cot (x)

Add a constant to the solution

= - 9 cot (x) + C

\int \frac{5 x ^{4}}{( 10 + x ^{5} ) ^{4}} d x

\int \frac{sin ( x )}{sec ( x )} d x

\int_{0}^{2} \frac{1}{( x - 2 ) ^{2}} d x

\int - \frac{y}{( x + 2 ) ^{2}} d x

f (x) = x^{3} + 3 x, at x = 2

What is the integral of (18)/(1-cos(2x)) ?
The integral of (18)/(1-cos(2x)) is -9cot(x)+C