What is the integral from 0 to 1 of cos(x^3) ?

The integral from 0 to 1 of cos(x^3) is

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integral from 0 to 1 of cos(x^3)

\int_{0}^{1} cos (x^{3}) d x

- \frac{1 \cdot ( 3 i 1 ^{3} Γ ( \frac{1}{3} , - i 1 ^{3} ) + 3 - i 1 ^{3} Γ ( \frac{1}{3} , i 1 ^{3} ) )}{6 3 1 ^{6}} - - \frac{0 \cdot ( 3 i 0 ^{3} Γ ( \frac{1}{3} , - i 0 ^{3} ) + 3 - i 0 ^{3} Γ ( \frac{1}{3} , i 0 ^{3} ) )}{6 3 0 ^{6}}

Solution steps

\int_{0}^{1} cos (x^{3}) d x

Use the nonelementary integral:

\int cos (x^{3}) d x = - \frac{x ( 3 i x ^{3} Γ ( \frac{1}{3} , - i x ^{3} ) + 3 - i x ^{3} Γ ( \frac{1}{3} , i x ^{3} ) )}{6 3 x ^{6}}

= [- \frac{x ( 3 i x ^{3} Γ ( \frac{1}{3} , - i x ^{3} ) + 3 - i x ^{3} Γ ( \frac{1}{3} , i x ^{3} ) )}{6 3 x ^{6}}]_{0}^{1}

Compute the boundaries:

- \frac{1 \cdot ( 3 i 1 ^{3} Γ ( \frac{1}{3} , - i 1 ^{3} ) + 3 - i 1 ^{3} Γ ( \frac{1}{3} , i 1 ^{3} ) )}{6 3 1 ^{6}} - - \frac{0 \cdot ( 3 i 0 ^{3} Γ ( \frac{1}{3} , - i 0 ^{3} ) + 3 - i 0 ^{3} Γ ( \frac{1}{3} , i 0 ^{3} ) )}{6 3 0 ^{6}}

= - \frac{1 \cdot ( 3 i 1 ^{3} Γ ( \frac{1}{3} , - i 1 ^{3} ) + 3 - i 1 ^{3} Γ ( \frac{1}{3} , i 1 ^{3} ) )}{6 3 1 ^{6}} - - \frac{0 \cdot ( 3 i 0 ^{3} Γ ( \frac{1}{3} , - i 0 ^{3} ) + 3 - i 0 ^{3} Γ ( \frac{1}{3} , i 0 ^{3} ) )}{6 3 0 ^{6}}

\int - 5 sin (5 x) d x

f (y) = 4 y^{2}

\frac{d}{d x} (x lo g_{2} (x - 3))

(ln (t))^{^{'}}

\frac{3 x - 5}{x ^{2} - 1}

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