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

The integral of sec^3(3x)tan(3x) is 1/9 sec^3(3x)+C

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integral of sec^3(3x)tan(3x)

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

\frac{1}{9} sec^{3} (3 x) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{3} d v

= \frac{1}{3} \cdot \int \frac{1}{3} d v

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot \frac{1}{3} v

Substitute back

= \frac{1}{3} \cdot \frac{1}{3} sec^{3} (3 x)

Simplify

\frac{1}{3} \cdot \frac{1}{3} sec^{3} (3 x) : \frac{1}{9} sec^{3} (3 x)

= \frac{1}{9} sec^{3} (3 x)

Add a constant to the solution

= \frac{1}{9} sec^{3} (3 x) + C

\frac{d}{d x} (\frac{9}{2 x ^{2}})

\frac{d y}{d x} = 20 - y

y^{^{'}} = \frac{( t ^{2} + t y + y ^{2} )}{t ^{2}}

h \to 0 lim (48 + 12 h + h^{2})

y = \frac{2 x - 1}{3 x + 5}

What is the integral of sec^3(3x)tan(3x) ?
The integral of sec^3(3x)tan(3x) is 1/9 sec^3(3x)+C