What is the integral of sec^3(4x) ?

The integral of sec^3(4x) is 1/4 (1/2 sec(4x)tan(4x)+1/2 ln|tan(4x)+sec(4x)|)+C

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integral of sec^3(4x)

\int sec^{3} (4 x) d x

\frac{1}{4} (\frac{1}{2} sec (4 x) tan (4 x) + \frac{1}{2} ln ∣ tan (4 x) + sec (4 x) ∣) + C

Solution steps

\int sec^{3} (4 x) d x

Apply u-substitution

: \int sec^{3} (u) \frac{1}{4} d u

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

Take the constant out:

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

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

Apply Integral Reduction

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

Substitute back

u = 4 x

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (1 \frac{1}{x})

\frac{d}{d x} ((3 x + 1)^{6} 2 x - 5)

\int 2 x d x

\frac{d}{d r} (r^{3})

f (x) = arctan (4 x)

What is the integral of sec^3(4x) ?
The integral of sec^3(4x) is 1/4 (1/2 sec(4x)tan(4x)+1/2 ln|tan(4x)+sec(4x)|)+C