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

The integral of 1/(2sin(2x)) is 1/4 ln|tan(x)|+C

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integral of 1/(2sin(2x))

integral

\frac{1}{2 sin ( 2 x )}

\frac{1}{4} ln ∣ tan (x) ∣ + C

Solution steps

\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

Rewrite using trig identities

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

Rewrite using trig identities

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

Apply Trigonometric Substitution

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

= \frac{1}{2} \cdot \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} \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = tan (x)

= \frac{1}{2} \cdot \frac{1}{2} ln ∣ tan (x) ∣

Simplify

\frac{1}{2} \cdot \frac{1}{2} ln ∣ tan (x) ∣ : \frac{1}{4} ln ∣ tan (x) ∣

= \frac{1}{4} ln ∣ tan (x) ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ tan (x) ∣ + C

(a^{2} + x^{2})^{\frac{1}{2}}

x^{4} (1 - x)^{5}

\frac{x ^{3}}{( x ^{2} + x + 2 )}

(x^{2} + 1)^{3}

ln (x - 1)

What is the integral of 1/(2sin(2x)) ?
The integral of 1/(2sin(2x)) is 1/4 ln|tan(x)|+C