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

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

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integral of sec^3(6x)

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

Apply Integral Reduction

= \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) ∣

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

Substitute back

u = 6 x

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

Add a constant to the solution

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

- x, x, - 2 x + 6

\int \frac{x - x ^{2}}{2 3 x} d x

\frac{\partial}{\partial x} (- 5 e^{x} ln (y))

y = x^{2}, y = 3 x + 4

- 6

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