What is the integral of 1/(sin(4x)) ?

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

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integral of 1/(sin(4x))

\int \frac{1}{sin ( 4 x )} d x

\frac{1}{4} ln ∣ tan (2 x) ∣ + C

Solution steps

\int \frac{1}{sin ( 4 x )} d x

Rewrite using trig identities

= \int csc (4 x) d x

Apply u-substitution

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

= \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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{4} ln ∣ v ∣

Substitute back

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

Simplify

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

= \frac{1}{4} ln ∣ tan (2 x) ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ tan (2 x) ∣ + C

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