What is the integral from 0 to x^3 of sin(t) ?

The integral from 0 to x^3 of sin(t) is -cos(x^3)+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to x^3 of sin(t)

\int_{0}^{x^{3}} sin (t) d t

- cos (x^{3}) + 1

Solution steps

\int_{0}^{x^{3}} sin (t) d t

Use the common integral:

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

= [- cos (t)]_{0}^{x^{3}}

Compute the boundaries:

- cos (x^{3}) + 1

= - cos (x^{3}) + 1

x \to - 5 + lim (ln (x + 5))

\frac{s + 5}{( s + 3 ) ( s - 1 )}

x \to 9 lim (\frac{- 1}{x - 9})

\int x \cdot 5^{x^{2}} d x

\int 10 cot^{3} (x) csc^{3} (x) d x

What is the integral from 0 to x^3 of sin(t) ?
The integral from 0 to x^3 of sin(t) is -cos(x^3)+1