What is the integral of tan(3x+1) ?

The integral of tan(3x+1) is -1/3 ln|cos(3x+1)|+C

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integral of tan(3x+1)

\int tan (3 x + 1) d x

- \frac{1}{3} ln ∣ cos (3 x + 1) ∣ + C

Solution steps

\int tan (3 x + 1) d x

Apply u-substitution

= \int tan (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) d u

Rewrite using trig identities

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

Apply u-substitution

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

Take the constant out:

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

= \frac{1}{3} (- \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 ∣ cos (3 x + 1) ∣)

Simplify

= - \frac{1}{3} ln ∣ cos (3 x + 1) ∣

Add a constant to the solution

= - \frac{1}{3} ln ∣ cos (3 x + 1) ∣ + C

What is the integral of tan(3x+1) ?
The integral of tan(3x+1) is -1/3 ln|cos(3x+1)|+C