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

The integral of xcos(4x^2) is 1/8 sin(4x^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of xcos(4x^2)

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

\frac{1}{8} sin (4 x^{2}) + C

Solution steps

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

Apply u-substitution

: \int \frac{cos ( u )}{8} d u

= \int \frac{cos ( u )}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{8} sin (u)

Substitute back

u = 4 x^{2}

= \frac{1}{8} sin (4 x^{2})

Add a constant to the solution

= \frac{1}{8} sin (4 x^{2}) + C

\frac{\partial}{\partial x} (sin^{2} (x - 3 y))

arctan (u)

\int_{0}^{π} sin^{2} (x) d x

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

\int 7 x^{- 3} d x

What is the integral of xcos(4x^2) ?
The integral of xcos(4x^2) is 1/8 sin(4x^2)+C