What is the integral of 3t^2sin(2t^3+1) ?

The integral of 3t^2sin(2t^3+1) is -1/2 cos(2t^3+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3t^2sin(2t^3+1)

\int 3 t^{2} sin (2 t^{3} + 1) d t

- \frac{1}{2} cos (2 t^{3} + 1) + C

Solution steps

\int 3 t^{2} sin (2 t^{3} + 1) d t

Take the constant out:

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

= 3 \cdot \int t^{2} sin (2 t^{3} + 1) d t

Apply u-substitution

: \int \frac{u sin ( 2 u ^{\frac{3}{2}} + 1 )}{2 t} d u

= 3 \cdot \int \frac{sin ( 2 u + 1 )}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int sin (2 u + 1) d u

Apply u-substitution

: \int sin (v) \frac{1}{2} d v

= 3 \cdot \frac{1}{3} \cdot \int sin (v) \frac{1}{2} d v

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \int sin (v) d v

Use the common integral:

\int sin (v) d v = - cos (v)

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

Substitute back

= 3 \cdot \frac{1}{3} \cdot \frac{1}{2} (- cos (2 t^{3} + 1))

Simplify

3 \cdot \frac{1}{3} \cdot \frac{1}{2} (- cos (2 t^{3} + 1)) : - \frac{1}{2} cos (2 t^{3} + 1)

= - \frac{1}{2} cos (2 t^{3} + 1)

Add a constant to the solution

= - \frac{1}{2} cos (2 t^{3} + 1) + C

\int \frac{2 x}{x} d x

n = 1 \sum \infty \frac{17 n !}{n ^{n}}

y^{2} = 4 x, y = 0, y = 4

(3 x^{2} + y + 3 x^{3} y) d x + (x) d y = 0

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

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