What is the integral of 3sec^2(3x) ?

The integral of 3sec^2(3x) is tan(3x)+C

integral of 3sec^2(3x)

\int 3 sec^{2} (3 x) d x

tan (3 x) + C

Solution steps

\int 3 sec^{2} (3 x) d x

Take the constant out:

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

= 3 \cdot \int sec^{2} (3 x) d x

Apply u-substitution

= 3 \cdot \int sec^{2} (u) \frac{1}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int sec^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= 3 \cdot \frac{1}{3} tan (u)

Substitute back

u = 3 x

= 3 \cdot \frac{1}{3} tan (3 x)

Simplify

3 \cdot \frac{1}{3} tan (3 x) : tan (3 x)

= tan (3 x)

Add a constant to the solution

= tan (3 x) + C

\frac{d}{d x} ((x^{2})^{3})

\frac{d}{d x} (\frac{sec ( x )}{1 + tan ( x )})

\frac{d}{d x} (- 0.0006 x (x - 15)^{2} (x - 40))

d er i v a t i v e y = x (7 x + 3)^{\frac{1}{3}}

x \to + 0 lim (x^{3} cos (\frac{π}{x}))