What is the integral of 2cos^2(4x)sin(4x) ?

The integral of 2cos^2(4x)sin(4x) is -1/6 cos^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 2cos^2(4x)sin(4x)

\int 2 cos^{2} (4 x) sin (4 x) d x

- \frac{1}{6} cos^{3} (4 x) + C

Solution steps

\int 2 cos^{2} (4 x) sin (4 x) d x

Take the constant out:

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

= 2 \cdot \int cos^{2} (4 x) sin (4 x) d x

Apply u-substitution

: \int cos^{2} (u) sin (u) \frac{1}{4} d u

= 2 \cdot \int cos^{2} (u) sin (u) \frac{1}{4} d u

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int cos^{2} (u) sin (u) d u

Apply u-substitution

: \int - v^{2} d v

= 2 \cdot \frac{1}{4} \cdot \int - v^{2} d v

Take the constant out:

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

= 2 \cdot \frac{1}{4} (- \int v^{2} d v)

Apply the Power Rule

: \frac{v ^{3}}{3}

= 2 \cdot \frac{1}{4} (- \frac{v ^{3}}{3})

Substitute back

= 2 \cdot \frac{1}{4} (- \frac{cos ^{3} ( 4 x )}{3})

Simplify

2 \cdot \frac{1}{4} (- \frac{cos ^{3} ( 4 x )}{3}) : - \frac{1}{6} cos^{3} (4 x)

= - \frac{1}{6} cos^{3} (4 x)

Add a constant to the solution

= - \frac{1}{6} cos^{3} (4 x) + C

\frac{\partial}{\partial x} ((x^{2} + y^{2} + z^{2})^{\frac{1}{2}})

x \to \infty lim (sec (x))

f (x) = 5 x, at x = 4, x = 9

x \to - 3 - lim (\frac{3 - x}{9 - x ^{2}})

What is the integral of 2cos^2(4x)sin(4x) ?
The integral of 2cos^2(4x)sin(4x) is -1/6 cos^3(4x)+C