What is the integral of sec^6(2x) ?

The integral of sec^6(2x) is 1/2 tan(2x)+1/3 tan^3(2x)+1/10 tan^5(2x)+C

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integral of sec^6(2x)

\int sec^{6} (2 x) d x

\frac{1}{2} tan (2 x) + \frac{1}{3} tan^{3} (2 x) + \frac{1}{10} tan^{5} (2 x) + C

Solution steps

\int sec^{6} (2 x) d x

Apply u-substitution

: \int sec^{6} (u) \frac{1}{2} d u

= \int sec^{6} (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int sec^{6} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

Substitute back

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

Simplify

\frac{1}{2} (tan (2 x) + \frac{2 tan ^{3} ( 2 x )}{3} + \frac{tan ^{5} ( 2 x )}{5}) : \frac{1}{2} tan (2 x) + \frac{1}{3} tan^{3} (2 x) + \frac{1}{10} tan^{5} (2 x)

= \frac{1}{2} tan (2 x) + \frac{1}{3} tan^{3} (2 x) + \frac{1}{10} tan^{5} (2 x)

Add a constant to the solution

= \frac{1}{2} tan (2 x) + \frac{1}{3} tan^{3} (2 x) + \frac{1}{10} tan^{5} (2 x) + C

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What is the integral of sec^6(2x) ?
The integral of sec^6(2x) is 1/2 tan(2x)+1/3 tan^3(2x)+1/10 tan^5(2x)+C