What is the integral of sec^3(1-4x) ?

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

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integral of sec^3(1-4x)

\int sec^{3} (1 - 4 x) d x

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

Solution steps

\int sec^{3} (1 - 4 x) d x

Apply u-substitution

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

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

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

Add a constant to the solution

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

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 0, y^{^{'}} (0) = 3

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

\int \frac{x ^{2}}{x - 7} d x

\frac{d}{d x} (\frac{1}{x ^{4}} - \frac{3}{x ^{2}})

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

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