What is the integral of sec^6(4x) ?

The integral of sec^6(4x) is 1/4 (tan(4x)+(2tan^3(4x))/3+(tan^5(4x))/5)+C

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integral of sec^6(4x)

\int sec^{6} (4 x) d x

\frac{1}{4} (tan (4 x) + \frac{2 tan ^{3} ( 4 x )}{3} + \frac{tan ^{5} ( 4 x )}{5}) + C

Solution steps

\int sec^{6} (4 x) d x

Apply u-substitution

: \int sec^{6} (u) \frac{1}{4} d u

= \int sec^{6} (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^{6} (u) d u

Rewrite using trig identities

= \frac{1}{4} \cdot \int sec^{2} (u) (1 + tan^{2} (u))^{2} d u

Apply u-substitution

: \int (1 + v^{2})^{2} d v

= \frac{1}{4} \cdot \int (1 + v^{2})^{2} d v

Expand

(1 + v^{2})^{2} : 1 + 2 v^{2} + v^{4}

= \frac{1}{4} \cdot \int 1 + 2 v^{2} + v^{4} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (\int 1 d v + \int 2 v^{2} d v + \int v^{4} d v)

\int 1 d v = v

\int 2 v^{2} d v = \frac{2 v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{4} (v + \frac{2 v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= \frac{1}{4} (tan (4 x) + \frac{2 tan ^{3} ( 4 x )}{3} + \frac{tan ^{5} ( 4 x )}{5})

Add a constant to the solution

= \frac{1}{4} (tan (4 x) + \frac{2 tan ^{3} ( 4 x )}{3} + \frac{tan ^{5} ( 4 x )}{5}) + C

f (x) = (3 x^{3} - 4 x^{2} + 5 x - 9)^{2}, at x = 2

x \to 2 lim (x^{2} + 6 x + 5)

x \to 0 + lim (x^{13 x})

x \to 4 lim (6)

\frac{\partial}{\partial y} (2 y - \frac{3}{y ^{2}} + x)

What is the integral of sec^6(4x) ?
The integral of sec^6(4x) is 1/4 (tan(4x)+(2tan^3(4x))/3+(tan^5(4x))/5)+C