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

The integral of sec^4(4x)tan(4x) is 1/16 sec^4(4x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{16} sec^{4} (4 x) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int v^{3} d v

= \frac{1}{4} \cdot \int v^{3} d v

Apply the Power Rule

: \frac{v ^{4}}{4}

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

Substitute back

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

Simplify

\frac{1}{4} \cdot \frac{sec ^{4} ( 4 x )}{4} : \frac{1}{16} sec^{4} (4 x)

= \frac{1}{16} sec^{4} (4 x)

Add a constant to the solution

= \frac{1}{16} sec^{4} (4 x) + C

\int 17 z^{3} e^{z} d z

\int x e^{\frac{x}{4}} d x

\int \frac{3 x ^{2}}{4 x - x ^{2}} d x

f (x) = x^{2} + 2

\frac{d y}{d x} = sin (3 x)

What is the integral of sec^4(4x)tan(4x) ?
The integral of sec^4(4x)tan(4x) is 1/16 sec^4(4x)+C