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

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

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integral of 8sec^3(3x)

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

\frac{8}{3} (\frac{1}{2} sec^{2} (3 x) sin (3 x) + \frac{1}{2} ln ∣ tan (3 x) + sec (3 x) ∣) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

= 8 \cdot \frac{1}{3} (\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}{3} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = 3 x

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

Simplify

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

= \frac{8}{3} (\frac{1}{2} sec^{2} (3 x) sin (3 x) + \frac{1}{2} ln ∣ tan (3 x) + sec (3 x) ∣)

Add a constant to the solution

= \frac{8}{3} (\frac{1}{2} sec^{2} (3 x) sin (3 x) + \frac{1}{2} ln ∣ tan (3 x) + sec (3 x) ∣) + C

y^{^{'}} + y \cdot cos (t) = sin (t) \cdot cos (t)

\int + \frac{x + 7}{x ^{2} + 4} d x

x^{3} (4 - x)

y^{^{'}} = \frac{x ^{2} - y ^{2}}{x y}, y (2) = 2

4^{6 x + 7}

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