What is the integral of (sec(3y)tan(36)) ?

The integral of (sec(3y)tan(36)) is tan(36) 1/3 ln|tan(3y)+sec(3y)|+C

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integral of (sec(3y)tan(36))

\int (sec (3 y) tan (36)) d y

tan (36) \frac{1}{3} ln ∣ tan (3 y) + sec (3 y) ∣ + C

Solution steps

\int sec (3 y) tan (36) d y

Take the constant out:

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

= tan (36) \cdot \int sec (3 y) d y

Apply u-substitution

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

= tan (36) \cdot \int sec (u) \frac{1}{3} d u

Take the constant out:

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

= tan (36) \frac{1}{3} \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= tan (36) \frac{1}{3} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = 3 y

= tan (36) \frac{1}{3} ln ∣ tan (3 y) + sec (3 y) ∣

Add a constant to the solution

= tan (36) \frac{1}{3} ln ∣ tan (3 y) + sec (3 y) ∣ + C

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What is the integral of (sec(3y)tan(36)) ?
The integral of (sec(3y)tan(36)) is tan(36) 1/3 ln|tan(3y)+sec(3y)|+C