What is the integral of sin(t^4) ?

The integral of sin(t^4) is 1/8 it(\E_{3/4}(-it^4)-\E_{3/4}(it^4))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(t^4)

\int sin (t^{4}) d t

\frac{1}{8} i t (E_{\frac{3}{4}} (- i t^{4}) - E_{\frac{3}{4}} (i t^{4})) + C

Solution steps

\int sin (t^{4}) d t

Use the nonelementary integral:

\int sin (x^{4}) d x = \frac{1}{8} i x (E_{\frac{3}{4}} (- i x^{4}) - E_{\frac{3}{4}} (i x^{4}))

= \frac{1}{8} i t (E_{\frac{3}{4}} (- i t^{4}) - E_{\frac{3}{4}} (i t^{4}))

Add a constant to the solution

= \frac{1}{8} i t (E_{\frac{3}{4}} (- i t^{4}) - E_{\frac{3}{4}} (i t^{4})) + C

f (t) = 1 + t

\frac{d}{d x} (y (x))

y^{^{'}} + \frac{t y}{t ^{2}} = \frac{4}{t ^{2}}

\frac{d y}{d x} = \frac{x}{7 cos ( y )}

\frac{d}{d x} (\frac{( x + 3 ) ^{2}}{( x + 2 ) ^{5} ( x + 4 ) ^{7}})

What is the integral of sin(t^4) ?
The integral of sin(t^4) is 1/8 it(\E_{3/4}(-it^4)-\E_{3/4}(it^4))+C