What is the integral from 2 to x^2 of sin(x^3) ?

The integral from 2 to x^2 of sin(x^3) is

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integral from 2 to x^2 of sin(x^3)

\int_{2}^{x^{2}} sin (x^{3}) d x

- \frac{1}{2} i \frac{x ^{2} Γ ( \frac{1}{3} , i ( x ^{2} ) ^{3} )}{3 3 i ( x ^{2} ) ^{3}} - \frac{x ^{2} Γ ( \frac{1}{3} , - i ( x ^{2} ) ^{3} )}{3 3 - i ( x ^{2} ) ^{3}} - (- \frac{1}{2} i (\frac{2Γ ( \frac{1}{3} , i 2 ^{3} )}{3 3 i 2 ^{3}} - \frac{2Γ ( \frac{1}{3} , - i 2 ^{3} )}{3 3 - i 2 ^{3}}))

Solution steps

\int_{2}^{x^{2}} sin (x^{3}) d x

Use the nonelementary integral:

\int sin (x^{3}) d x = - \frac{1}{2} i (\frac{x Γ ( \frac{1}{3} , i x ^{3} )}{3 3 i x ^{3}} - \frac{x Γ ( \frac{1}{3} , - i x ^{3} )}{3 3 - i x ^{3}})

= [- \frac{1}{2} i (\frac{x Γ ( \frac{1}{3} , i x ^{3} )}{3 3 i x ^{3}} - \frac{x Γ ( \frac{1}{3} , - i x ^{3} )}{3 3 - i x ^{3}})]_{2}^{x^{2}}

Compute the boundaries:

- \frac{1}{2} i \frac{x ^{2} Γ ( \frac{1}{3} , i ( x ^{2} ) ^{3} )}{3 3 i ( x ^{2} ) ^{3}} - \frac{x ^{2} Γ ( \frac{1}{3} , - i ( x ^{2} ) ^{3} )}{3 3 - i ( x ^{2} ) ^{3}} - (- \frac{1}{2} i (\frac{2Γ ( \frac{1}{3} , i 2 ^{3} )}{3 3 i 2 ^{3}} - \frac{2Γ ( \frac{1}{3} , - i 2 ^{3} )}{3 3 - i 2 ^{3}}))

= - \frac{1}{2} i \frac{x ^{2} Γ ( \frac{1}{3} , i ( x ^{2} ) ^{3} )}{3 3 i ( x ^{2} ) ^{3}} - \frac{x ^{2} Γ ( \frac{1}{3} , - i ( x ^{2} ) ^{3} )}{3 3 - i ( x ^{2} ) ^{3}} - (- \frac{1}{2} i (\frac{2Γ ( \frac{1}{3} , i 2 ^{3} )}{3 3 i 2 ^{3}} - \frac{2Γ ( \frac{1}{3} , - i 2 ^{3} )}{3 3 - i 2 ^{3}}))

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What is the integral from 2 to x^2 of sin(x^3) ?
The integral from 2 to x^2 of sin(x^3) is