What is the integral of tan(4x) ?

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

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integral of tan(4x)

\int tan (4 x) d x

- \frac{1}{4} ln ∣ cos (4 x) ∣ + C

Solution steps

\int tan (4 x) d x

Rewrite using trig identities

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

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{4} ln ∣ u ∣

Substitute back

u = cos (4 x)

= - \frac{1}{4} ln ∣ cos (4 x) ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ cos (4 x) ∣ + C

\int e^{- x} tan (e^{- x}) d x

\int \frac{1}{25 - 16 x ^{2}} d x

\int \frac{x + 3}{x ^{2} - 4 x + 3} d x

f (x) = 3 x - 2 x, (1, 1)

z \to + 10 lim (\frac{1}{z - 10})