What is the integral of sec^3(4x)tan(4x) ?

The integral of sec^3(4x)tan(4x) is 1/12 sec^3(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^3(4x)tan(4x)

\int sec^{3} (4 x) tan (4 x) d x

\frac{1}{12} sec^{3} (4 x) + C

Solution steps

\int sec^{3} (4 x) tan (4 x) d x

Apply u-substitution

: \int sec^{3} (u) tan (u) \frac{1}{4} d u

= \int sec^{3} (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^{3} (u) tan (u) d u

Apply u-substitution

: \int \frac{1}{3} d v

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

Integral of a constant:

\int a d x = a x

= \frac{1}{4} \cdot \frac{1}{3} v

Substitute back

= \frac{1}{4} \cdot \frac{1}{3} sec^{3} (4 x)

Simplify

\frac{1}{4} \cdot \frac{1}{3} sec^{3} (4 x) : \frac{1}{12} sec^{3} (4 x)

= \frac{1}{12} sec^{3} (4 x)

Add a constant to the solution

= \frac{1}{12} sec^{3} (4 x) + C

\int \frac{1}{5 x} d x

\int_{0}^{2} 2 x^{2} e^{- 2 x} d x

\frac{d}{d x} (\frac{( 2 x ^{2} + 1 ) ^{3}}{x ^{2} - 1})

\int \frac{1}{( 1 + 9 x ^{2} ) ^{\frac{3}{2}}} d x

\int_{1}^{4} - 3 e^{x} d x

What is the integral of sec^3(4x)tan(4x) ?
The integral of sec^3(4x)tan(4x) is 1/12 sec^3(4x)+C