What is the integral of-3tan(3y) ?

The integral of-3tan(3y) is ln|cos(3y)|+C

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integral of-3tan(3y)

\int - 3 tan (3 y) d y

ln ∣ cos (3 y) ∣ + C

Solution steps

\int - 3 tan (3 y) d y

Take the constant out:

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

= - 3 \cdot \int tan (3 y) d y

Rewrite using trig identities

= - 3 \cdot \int \frac{sin ( 3 y )}{cos ( 3 y )} d y

Apply u-substitution

: \int - \frac{1}{3 u} d u

= - 3 \cdot \int - \frac{1}{3 u} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 3 (- \frac{1}{3} ln ∣ u ∣)

Substitute back

u = cos (3 y)

= - 3 (- \frac{1}{3} ln ∣ cos (3 y) ∣)

Simplify

- 3 (- \frac{1}{3} ln ∣ cos (3 y) ∣) : ln ∣ cos (3 y) ∣

= ln ∣ cos (3 y) ∣

Add a constant to the solution

= ln ∣ cos (3 y) ∣ + C

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