What is the integral of 8tan(4x+1) ?

The integral of 8tan(4x+1) is -2ln|cos(4x+1)|+C

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integral of 8tan(4x+1)

\int 8 tan (4 x + 1) d x

- 2 ln ∣ cos (4 x + 1) ∣ + C

Solution steps

\int 8 tan (4 x + 1) d x

Take the constant out:

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

= 8 \cdot \int tan (4 x + 1) d x

Apply u-substitution

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

= 8 \cdot \int tan (u) \frac{1}{4} d u

Take the constant out:

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

= 8 \cdot \frac{1}{4} \cdot \int tan (u) d u

Rewrite using trig identities

= 8 \cdot \frac{1}{4} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

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

= 8 \cdot \frac{1}{4} \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

= 8 \cdot \frac{1}{4} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 8 \cdot \frac{1}{4} (- ln ∣ v ∣)

Substitute back

= 8 \cdot \frac{1}{4} (- ln ∣ cos (4 x + 1) ∣)

Simplify

8 \cdot \frac{1}{4} (- ln ∣ cos (4 x + 1) ∣) : - 2 ln ∣ cos (4 x + 1) ∣

= - 2 ln ∣ cos (4 x + 1) ∣

Add a constant to the solution

= - 2 ln ∣ cos (4 x + 1) ∣ + C

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What is the integral of 8tan(4x+1) ?
The integral of 8tan(4x+1) is -2ln|cos(4x+1)|+C