What is the integral of tan(3x)cot(3x)sec(3x)csc(3x) ?

The integral of tan(3x)cot(3x)sec(3x)csc(3x) is 1/3 ln|tan(3x)|+C

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integral of tan(3x)cot(3x)sec(3x)csc(3x)

\int tan (3 x) cot (3 x) sec (3 x) csc (3 x) d x

\frac{1}{3} ln ∣ tan (3 x) ∣ + C

Solution steps

\int tan (3 x) cot (3 x) sec (3 x) csc (3 x) d x

Apply u-substitution

: \int tan (u) cot (u) sec (u) csc (u) \frac{1}{3} d u

= \int tan (u) cot (u) sec (u) csc (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int tan (u) cot (u) sec (u) csc (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int \frac{sec ( u )}{sin ( u )} d u

Apply Trigonometric Substitution

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

= \frac{1}{3} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{3} ln ∣ v ∣

Substitute back

= \frac{1}{3} ln ∣ tan (3 x) ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ tan (3 x) ∣ + C

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What is the integral of tan(3x)cot(3x)sec(3x)csc(3x) ?
The integral of tan(3x)cot(3x)sec(3x)csc(3x) is 1/3 ln|tan(3x)|+C