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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Simplify

sec (4 x) tan (4 x) sec^{4} (4 x) : sec^{5} (4 x) tan (4 x)

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

Apply u-substitution

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

= 4 \cdot \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

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

Apply u-substitution

: \int v^{4} d v

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

Apply the Power Rule

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

e^{- 4 x^{2}}

\frac{2}{( 1 + x ) ^{2}}

\int x e^{i x} d x

f (x) = 4 x^{3} - 24 x^{2}, at x = - 3

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

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