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

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

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integral of-8tan(4x)

\int - 8 tan (4 x) d x

2 ln ∣ cos (4 x) ∣ + C

Solution steps

\int - 8 tan (4 x) 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) d x

Rewrite using trig identities

= - 8 \cdot \int \frac{sin ( 4 x )}{cos ( 4 x )} d x

Apply u-substitution

: \int - \frac{1}{4 u} d u

= - 8 \cdot \int - \frac{1}{4 u} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 8 (- \frac{1}{4} ln ∣ u ∣)

Substitute back

u = cos (4 x)

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

Simplify

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

= 2 ln ∣ cos (4 x) ∣

Add a constant to the solution

= 2 ln ∣ cos (4 x) ∣ + C

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