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

The integral of 1/(tan(2x)) is 1/2 ln|sin(2x)|+C

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integral of 1/(tan(2x))

\int \frac{1}{tan ( 2 x )} d x

\frac{1}{2} ln ∣ sin (2 x) ∣ + C

Solution steps

\int \frac{1}{tan ( 2 x )} d x

Rewrite using trig identities

= \int \frac{cos ( 2 x )}{sin ( 2 x )} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

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}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = sin (2 x)

= \frac{1}{2} ln ∣ sin (2 x) ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ sin (2 x) ∣ + C

(9 - x)^{\frac{1}{2}}

y = \frac{1}{3 x ^{2}} - \frac{5}{2 x}

f (x) = \frac{2 - ln ( x )}{2 x x}

y^{^{'''}} - 2 y^{^{''}} - 4 y^{^{'}} + 8 y = 0

\frac{\partial}{\partial y} (- 4 x^{3} + 4 x)

What is the integral of 1/(tan(2x)) ?
The integral of 1/(tan(2x)) is 1/2 ln|sin(2x)|+C