What is the integral from 0 to x of cos(t^4) ?

The integral from 0 to x of cos(t^4) is -(x(\E_{3/4}(ix^4)+\E_{3/4}(-ix^4)))/8

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integral from 0 to x of cos(t^4)

\int_{0}^{x} cos (t^{4}) d t

- \frac{x ( E _{\frac{3}{4}} ( i x ^{4} ) + E _{\frac{3}{4}} ( - i x ^{4} ) )}{8}

Solution steps

\int_{0}^{x} cos (t^{4}) d t

Use the nonelementary integral:

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

= [- \frac{1}{8} t (E_{\frac{3}{4}} (- i t^{4}) + E_{\frac{3}{4}} (i t^{4}))]_{0}^{x}

Compute the boundaries:

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

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

Simplify

= - \frac{x ( E _{\frac{3}{4}} ( i x ^{4} ) + E _{\frac{3}{4}} ( - i x ^{4} ) )}{8}

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What is the integral from 0 to x of cos(t^4) ?
The integral from 0 to x of cos(t^4) is -(x(\E_{3/4}(ix^4)+\E_{3/4}(-ix^4)))/8