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

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

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integral of 1/(sin(2y))

\int \frac{1}{sin ( 2 y )} d y

\frac{1}{2} ln ∣ tan (y) ∣ + C

Solution steps

\int \frac{1}{sin ( 2 y )} d y

Rewrite using trig identities

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

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}{cos ( y ) sin ( y )} d y

Rewrite using trig identities

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

Apply Trigonometric Substitution

= \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 = tan (y)

= \frac{1}{2} ln ∣ tan (y) ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ tan (y) ∣ + C

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