What is the integral of 7sec^3(3x) ?

The integral of 7sec^3(3x) is 7/3 (1/2 sec^2(3x)sin(3x)+1/2 ln|tan(3x)+sec(3x)|)+C

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integral of 7sec^3(3x)

\int 7 sec^{3} (3 x) d x

\frac{7}{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 7 sec^{3} (3 x) d x

Take the constant out:

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

= 7 \cdot \int sec^{3} (3 x) d x

Apply u-substitution

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

= 7 \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

= 7 \cdot \frac{1}{3} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 7 \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) ∣

= 7 \cdot \frac{1}{3} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = 3 x

= 7 \cdot \frac{1}{3} (\frac{sec ^{2} ( 3 x ) sin ( 3 x )}{2} + \frac{1}{2} ln ∣ tan (3 x) + sec (3 x) ∣)

Simplify

7 \cdot \frac{1}{3} (\frac{sec ^{2} ( 3 x ) sin ( 3 x )}{2} + \frac{1}{2} ln ∣ tan (3 x) + sec (3 x) ∣) : \frac{7}{3} (\frac{1}{2} sec^{2} (3 x) sin (3 x) + \frac{1}{2} ln ∣ tan (3 x) + sec (3 x) ∣)

= \frac{7}{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{7}{3} (\frac{1}{2} sec^{2} (3 x) sin (3 x) + \frac{1}{2} ln ∣ tan (3 x) + sec (3 x) ∣) + C

\int (\frac{8}{3} + 4 u^{2}) d u

\frac{d ^{2}}{d x ^{2}} (1 + x)

\frac{\partial}{\partial y} (cos (8 x) sin (2 y))

x^{4} - 3 x^{2} + 1

What is the integral of 7sec^3(3x) ?
The integral of 7sec^3(3x) is 7/3 (1/2 sec^2(3x)sin(3x)+1/2 ln|tan(3x)+sec(3x)|)+C