What is the integral of sec^3(8x) ?

The integral of sec^3(8x) is 1/8 ((sec^2(8x)sin(8x))/2+1/2 ln|tan(8x)+sec(8x)|)+C

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integral of sec^3(8x)

\int sec^{3} (8 x) d x

\frac{1}{8} (\frac{sec ^{2} ( 8 x ) sin ( 8 x )}{2} + \frac{1}{2} ln ∣ tan (8 x) + sec (8 x) ∣) + C

Solution steps

\int sec^{3} (8 x) d x

Apply u-substitution

: \int sec^{3} (u) \frac{1}{8} d u

= \int sec^{3} (u) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{8} (\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}{8} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = 8 x

= \frac{1}{8} (\frac{sec ^{2} ( 8 x ) sin ( 8 x )}{2} + \frac{1}{2} ln ∣ tan (8 x) + sec (8 x) ∣)

Add a constant to the solution

= \frac{1}{8} (\frac{sec ^{2} ( 8 x ) sin ( 8 x )}{2} + \frac{1}{2} ln ∣ tan (8 x) + sec (8 x) ∣) + C

\int x^{2} 2 x + 5 d x

y = \frac{3}{x}, (4, \frac{3}{2})

t^{2} cos (t)

y^{^{''}} + 4 y^{^{'}} + 4 y = t + e^{2 t}

f (x) = 9 x^{2}

What is the integral of sec^3(8x) ?
The integral of sec^3(8x) is 1/8 ((sec^2(8x)sin(8x))/2+1/2 ln|tan(8x)+sec(8x)|)+C