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

The integral of sec^6(2x)tan(2x) is 1/12 sec^6(2x)+C

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integral of sec^6(2x)tan(2x)

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

\frac{1}{12} sec^{6} (2 x) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int v^{5} d v

= \frac{1}{2} \cdot \int v^{5} d v

Apply the Power Rule

: \frac{v ^{6}}{6}

= \frac{1}{2} \cdot \frac{v ^{6}}{6}

Substitute back

= \frac{1}{2} \cdot \frac{sec ^{6} ( 2 x )}{6}

Simplify

\frac{1}{2} \cdot \frac{sec ^{6} ( 2 x )}{6} : \frac{1}{12} sec^{6} (2 x)

= \frac{1}{12} sec^{6} (2 x)

Add a constant to the solution

= \frac{1}{12} sec^{6} (2 x) + C

cos (x) \frac{d y}{d x} - sin (x) y = 0

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

x \to \infty lim (ln (1))

y^{^{'}} = 8 x y - 2 x, y (0) = 16

\frac{\partial}{\partial x} (cot (3^{2}) + 5)

What is the integral of sec^6(2x)tan(2x) ?
The integral of sec^6(2x)tan(2x) is 1/12 sec^6(2x)+C