What is the integral of sec^5(4x)tan(4x) ?

The integral of sec^5(4x)tan(4x) is 1/20 sec^5(4x)+C

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integral of sec^5(4x)tan(4x)

\int sec^{5} (4 x) tan (4 x) d x

\frac{1}{20} sec^{5} (4 x) + C

Solution steps

\int sec^{5} (4 x) tan (4 x) d x

Apply u-substitution

: \int sec^{5} (u) tan (u) \frac{1}{4} d u

= \int sec^{5} (u) tan (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^{5} (u) tan (u) d u

Apply u-substitution

: \int v^{4} d v

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

Apply the Power Rule

: \frac{v ^{5}}{5}

= \frac{1}{4} \cdot \frac{v ^{5}}{5}

Substitute back

= \frac{1}{4} \cdot \frac{sec ^{5} ( 4 x )}{5}

Simplify

\frac{1}{4} \cdot \frac{sec ^{5} ( 4 x )}{5} : \frac{1}{20} sec^{5} (4 x)

= \frac{1}{20} sec^{5} (4 x)

Add a constant to the solution

= \frac{1}{20} sec^{5} (4 x) + C

x \to - 1 lim (10 x)

x y + 5 e^{y} = 5 e

\frac{d}{d x} (tan (a x + b))

\frac{\partial}{\partial x} (3 y^{2} x - y^{3} - x^{4})

x^{2} y^{^{''}} = y^{^{'}} + (y^{^{'}})^{2}

What is the integral of sec^5(4x)tan(4x) ?
The integral of sec^5(4x)tan(4x) is 1/20 sec^5(4x)+C