What is the integral of tan(4-9x) ?

The integral of tan(4-9x) is 1/9 ln|cos(4-9x)|+C

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integral of tan(4-9x)

\int tan (4 - 9 x) d x

\frac{1}{9} ln ∣ cos (4 - 9 x) ∣ + C

Solution steps

\int tan (4 - 9 x) d x

Rewrite using trig identities

= \int \frac{sin ( 4 - 9 x )}{cos ( 4 - 9 x )} d x

Apply u-substitution

= \int \frac{1}{9 u} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{9} ln ∣ u ∣

Substitute back

u = cos (4 - 9 x)

= \frac{1}{9} ln ∣ cos (4 - 9 x) ∣

Add a constant to the solution

= \frac{1}{9} ln ∣ cos (4 - 9 x) ∣ + C

What is the integral of tan(4-9x) ?
The integral of tan(4-9x) is 1/9 ln|cos(4-9x)|+C