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

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

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integral of 9tan(9x)

\int 9 tan (9 x) d x

- ln ∣ cos (9 x) ∣ + C

Solution steps

\int 9 tan (9 x) d x

Take the constant out:

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

= 9 \cdot \int tan (9 x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int - \frac{1}{9 u} d u

= 9 \cdot \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

= 9 (- \frac{1}{9} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 9 (- \frac{1}{9} ln ∣ u ∣)

Substitute back

u = cos (9 x)

= 9 (- \frac{1}{9} ln ∣ cos (9 x) ∣)

Simplify

9 (- \frac{1}{9} ln ∣ cos (9 x) ∣) : - ln ∣ cos (9 x) ∣

= - ln ∣ cos (9 x) ∣

Add a constant to the solution

= - ln ∣ cos (9 x) ∣ + C

x \to 5 lim (4 x)

\int (3 x + 2 y) d x

\frac{d}{d x} (- x ln (x))

5 t^{\frac{7}{5}} + 5 e^{3}