What is the integral of sec^3(7x-1) ?

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

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integral of sec^3(7x-1)

\int sec^{3} (7 x - 1) d x

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

Solution steps

\int sec^{3} (7 x - 1) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

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

Substitute back

u = 7 x - 1

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

Add a constant to the solution

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

\int \frac{t ^{2}}{( t ^{2} - 9 ) ^{\frac{5}{2}}} d t

\frac{d}{d t} (6 t - t^{3})

(1 - x^{2})^{\frac{1}{2}}

x \to p (x) lim (\frac{f ( x ) - f ( p )}{x - p ( x )})

h \to 0 lim (\frac{sin ( \frac{π}{3} + h ) - sin ( \frac{π}{3} )}{h})

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