What is the integral from 0 to t of sin(u^2) ?

The integral from 0 to t of sin(u^2) is sqrt(pi/2)\S(sqrt(2/pi)t)-sqrt(pi/2)\S(sqrt(2/pi)*0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to t of sin(u^2)

\int_{0}^{t} sin (u^{2}) d u

\frac{π}{2} § (\frac{2}{π} t) - \frac{π}{2} § (\frac{2}{π} \cdot 0)

Solution steps

\int_{0}^{t} sin (u^{2}) d u

Use the nonelementary integral:

\int sin (x^{2}) d x = \frac{π}{2} § (\frac{2}{π} x)

= [\frac{π}{2} § (\frac{2}{π} u)]_{0}^{t}

Compute the boundaries:

\frac{π}{2} § (\frac{2}{π} t) - \frac{π}{2} § (\frac{2}{π} \cdot 0)

= \frac{π}{2} § (\frac{2}{π} t) - \frac{π}{2} § (\frac{2}{π} \cdot 0)

\int_{0}^{1} x 8^{x} d x

\int_{0}^{1} x^{5} 1 + 4 x^{2} d x

\int_{- 1}^{2} ∣ x ∣ - x d x

\int_{0}^{x} 7 e^{- 7 x} d x

\int_{1}^{1.5} x^{2} \cdot ln (x) d x

What is the integral from 0 to t of sin(u^2) ?
The integral from 0 to t of sin(u^2) is sqrt(pi/2)\S(sqrt(2/pi)t)-sqrt(pi/2)\S(sqrt(2/pi)*0)