What is the integral from 0 to 1 of sin(x^4) ?

The integral from 0 to 1 of sin(x^4) is i/8 (\E_{3/4}(-1i)-\E_{3/4}(1i))

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integral from 0 to 1 of sin(x^4)

\int_{0}^{1} sin (x^{4}) d x

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

Solution steps

\int_{0}^{1} sin (x^{4}) d x

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 x (E_{\frac{3}{4}} (- i x^{4}) - E_{\frac{3}{4}} (i x^{4}))]_{0}^{1}

Compute the boundaries:

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

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

Simplify

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

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What is the integral from 0 to 1 of sin(x^4) ?
The integral from 0 to 1 of sin(x^4) is i/8 (\E_{3/4}(-1i)-\E_{3/4}(1i))