What is the integral of 8sec^3(6x) ?

The integral of 8sec^3(6x) is 4/3 (1/2 sec^2(6x)sin(6x)+1/2 ln|tan(6x)+sec(6x)|)+C

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integral of 8sec^3(6x)

\int 8 sec^{3} (6 x) d x

\frac{4}{3} (\frac{1}{2} sec^{2} (6 x) sin (6 x) + \frac{1}{2} ln ∣ tan (6 x) + sec (6 x) ∣) + C

Solution steps

\int 8 sec^{3} (6 x) d x

Take the constant out:

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

= 8 \cdot \int sec^{3} (6 x) d x

Apply u-substitution

: \int sec^{3} (u) \frac{1}{6} d u

= 8 \cdot \int sec^{3} (u) \frac{1}{6} d u

Take the constant out:

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

= 8 \cdot \frac{1}{6} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 8 \cdot \frac{1}{6} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 8 \cdot \frac{1}{6} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = 6 x

= 8 \cdot \frac{1}{6} (\frac{sec ^{2} ( 6 x ) sin ( 6 x )}{2} + \frac{1}{2} ln ∣ tan (6 x) + sec (6 x) ∣)

Simplify

8 \cdot \frac{1}{6} (\frac{sec ^{2} ( 6 x ) sin ( 6 x )}{2} + \frac{1}{2} ln ∣ tan (6 x) + sec (6 x) ∣) : \frac{4}{3} (\frac{1}{2} sec^{2} (6 x) sin (6 x) + \frac{1}{2} ln ∣ tan (6 x) + sec (6 x) ∣)

= \frac{4}{3} (\frac{1}{2} sec^{2} (6 x) sin (6 x) + \frac{1}{2} ln ∣ tan (6 x) + sec (6 x) ∣)

Add a constant to the solution

= \frac{4}{3} (\frac{1}{2} sec^{2} (6 x) sin (6 x) + \frac{1}{2} ln ∣ tan (6 x) + sec (6 x) ∣) + C

\frac{\partial}{\partial x} (3 e^{2 x} - 5 ln (y) + 7)

\int_{2}^{7} x d x

\frac{d y}{d x} = (\frac{3 y + 2}{5 x + 4})^{2}

\int ln (2 x) d x

\frac{\partial}{\partial x} (sin (4 x))

What is the integral of 8sec^3(6x) ?
The integral of 8sec^3(6x) is 4/3 (1/2 sec^2(6x)sin(6x)+1/2 ln|tan(6x)+sec(6x)|)+C