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

The integral of 2cos(4x)e^{sin(4x)} is 1/2 e^{sin(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(4x)e^{sin(4x)}

\int 2 cos (4 x) e^{s i n (4 x)} d x

\frac{1}{2} e^{s i n (4 x)} + C

Solution steps

\int 2 cos (4 x) e^{s i n (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 (4 x) e^{s i n (4 x)} d x

Apply u-substitution

: \int cos (u) e^{s i n (u)} \frac{1}{4} d u

= 2 \cdot \int cos (u) e^{s i n (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 (u) e^{s i n (u)} d u

Apply u-substitution

: \int e^{v} d v

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

Use the common integral:

\int e^{v} d v = e^{v}

= 2 \cdot \frac{1}{4} e^{v}

Substitute back

= 2 \cdot \frac{1}{4} e^{s i n (4 x)}

Simplify

2 \cdot \frac{1}{4} e^{s i n (4 x)} : \frac{1}{2} e^{s i n (4 x)}

= \frac{1}{2} e^{s i n (4 x)}

Add a constant to the solution

= \frac{1}{2} e^{s i n (4 x)} + C

2 n d

f (x) = \frac{1 + 2 x}{3 + x}, at x = 2

\frac{\partial}{\partial x} (5 2 x + 2 y)

\frac{d}{d x} y = (5 - x)^{63}

y = 3 x^{3} - x^{2} + 4, (2, 24)

What is the integral of 2cos(4x)e^{sin(4x)} ?
The integral of 2cos(4x)e^{sin(4x)} is 1/2 e^{sin(4x)}+C