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

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

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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

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

= 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

x \to 1 lim (\frac{x}{ln ( x )})

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

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

x y^{^{''}} + 4 y^{^{'}} = 0

\int_{0}^{2} arctan (\frac{x}{2}) d x