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

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

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integral of sec^3(7x)

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

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

Solution steps

\int sec^{3} (7 x) 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

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

Add a constant to the solution

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

y^{^{'}} = sec (2) (x - y)

f (1) = 2 x^{4} - x^{3} + 6 x^{2}

x \to \infty lim (3 - x)

\frac{d}{d x} ((3 x - 7)^{12})

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