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

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

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integral of 1/(sin(2y)cos(2y))

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

\frac{1}{2} ln ∣ tan (2 y) ∣ + C

Solution steps

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

Rewrite using trig identities

= \int \frac{2}{sin ( 4 y )} d y

Take the constant out:

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

= 2 \cdot \int \frac{1}{sin ( 4 y )} d y

Rewrite using trig identities

= 2 \cdot \int csc (4 y) d y

Apply u-substitution

: \int csc (u) \frac{1}{4} d u

= 2 \cdot \int csc (u) \frac{1}{4} d u

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int csc (u) d u

Apply u-substitution

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

= 2 \cdot \frac{1}{4} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 \cdot \frac{1}{4} ln ∣ v ∣

Substitute back

= 2 \cdot \frac{1}{4} ln tan (\frac{4 y}{2})

Simplify

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

= \frac{1}{2} ln ∣ tan (2 y) ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ tan (2 y) ∣ + C

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