What is the integral of-2tcos(-t^2) ?

The integral of-2tcos(-t^2) is -sin(t^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-2tcos(-t^2)

\int - 2 t cos (- t^{2}) d t

- sin (t^{2}) + C

Solution steps

\int - 2 t cos (- t^{2}) d t

Take the constant out:

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

= - 2 \cdot \int t cos (- t^{2}) d t

Apply u-substitution

: \int \frac{cos ( u )}{2} d u

= - 2 \cdot \int \frac{cos ( u )}{2} d u

Take the constant out:

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

= - 2 \cdot \frac{1}{2} \cdot \int cos (u) d u

Use the common integral:

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

= - 2 \cdot \frac{1}{2} sin (u)

Substitute back

u = t^{2}

= - 2 \cdot \frac{1}{2} sin (t^{2})

Simplify

- 2 \cdot \frac{1}{2} sin (t^{2}) : - sin (t^{2})

= - sin (t^{2})

Add a constant to the solution

= - sin (t^{2}) + C

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

\frac{d y}{d x} + \frac{y}{x} = 8 x^{7} y^{2}

\frac{\partial}{\partial x} (4 y - 4 x^{3})

(1.5) (- 1.7)

\int 4 x y^{3} d x

What is the integral of-2tcos(-t^2) ?
The integral of-2tcos(-t^2) is -sin(t^2)+C