What is the integral of sec^3(9x) ?

The integral of sec^3(9x) is 1/9 ((sec^2(9x)sin(9x))/2+1/2 ln|tan(9x)+sec(9x)|)+C

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integral of sec^3(9x)

\int sec^{3} (9 x) d x

\frac{1}{9} (\frac{sec ^{2} ( 9 x ) sin ( 9 x )}{2} + \frac{1}{2} ln ∣ tan (9 x) + sec (9 x) ∣) + C

Solution steps

\int sec^{3} (9 x) d x

Apply u-substitution

: \int sec^{3} (u) \frac{1}{9} d u

= \int sec^{3} (u) \frac{1}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{9} (\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}{9} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = 9 x

= \frac{1}{9} (\frac{sec ^{2} ( 9 x ) sin ( 9 x )}{2} + \frac{1}{2} ln ∣ tan (9 x) + sec (9 x) ∣)

Add a constant to the solution

= \frac{1}{9} (\frac{sec ^{2} ( 9 x ) sin ( 9 x )}{2} + \frac{1}{2} ln ∣ tan (9 x) + sec (9 x) ∣) + C

f (x) = (ln (x))^{3}, at x = 3

\int \frac{( 5 x + 1 )}{( 2 x + 1 ) ( x - 1 )} d x

y^{^{''''}} + 3 y^{^{''}} - 4 y = 0

arctan (\frac{x - 1}{x + 1})

\int 2 e^{2 x - 1} d x

What is the integral of sec^3(9x) ?
The integral of sec^3(9x) is 1/9 ((sec^2(9x)sin(9x))/2+1/2 ln|tan(9x)+sec(9x)|)+C