What is the integral of sec^5(3x) ?

The integral of sec^5(3x) is 1/12 sec^4(3x)sin(3x)+1/8 (sec(3x)tan(3x)+ln|tan(3x)+sec(3x)|)+C

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integral of sec^5(3x)

\int sec^{5} (3 x) d x

\frac{1}{12} sec^{4} (3 x) sin (3 x) + \frac{1}{8} (sec (3 x) tan (3 x) + ln ∣ tan (3 x) + sec (3 x) ∣) + C

Solution steps

\int sec^{5} (3 x) d x

Apply u-substitution

: \int sec^{5} (u) \frac{1}{3} d u

= \int sec^{5} (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sec^{5} (u) d u

Apply Integral Reduction

= \frac{1}{3} (\frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} \cdot \int sec^{3} (u) d u)

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

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

Substitute back

u = 3 x

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

Simplify

\frac{1}{3} (\frac{sec ^{4} ( 3 x ) sin ( 3 x )}{4} + \frac{3}{4} (\frac{1}{2} sec (3 x) tan (3 x) + \frac{1}{2} ln ∣ tan (3 x) + sec (3 x) ∣)) : \frac{1}{12} sec^{4} (3 x) sin (3 x) + \frac{1}{8} (sec (3 x) tan (3 x) + ln ∣ tan (3 x) + sec (3 x) ∣)

= \frac{1}{12} sec^{4} (3 x) sin (3 x) + \frac{1}{8} (sec (3 x) tan (3 x) + ln ∣ tan (3 x) + sec (3 x) ∣)

Add a constant to the solution

= \frac{1}{12} sec^{4} (3 x) sin (3 x) + \frac{1}{8} (sec (3 x) tan (3 x) + ln ∣ tan (3 x) + sec (3 x) ∣) + C

\frac{d}{d x} (\frac{2 x y}{( 2 - x ^{2} )})

cos (6 y)

\frac{d}{d x} (\frac{a x + b}{c x + d})

x \to 1 lim (arccos (x))

\frac{d}{d s} (\frac{6}{s ^{2} + 36})

What is the integral of sec^5(3x) ?
The integral of sec^5(3x) is 1/12 sec^4(3x)sin(3x)+1/8 (sec(3x)tan(3x)+ln|tan(3x)+sec(3x)|)+C