What is the integral from \sqrt[3]{x} to pi of sin(t^3) ?

The integral from \sqrt[3]{x} to pi of sin(t^3) is

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integral from \sqrt[3]{x} to pi of sin(t^3)

\int_{3 x}^{π} sin (t^{3}) d t

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

Solution steps

\int_{3 x}^{π} sin (t^{3}) d t

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{t Γ ( \frac{1}{3} , i t ^{3} )}{3 3 i t ^{3}} - \frac{t Γ ( \frac{1}{3} , - i t ^{3} )}{3 3 - i t ^{3}})]_{3 x}^{π}

Compute the boundaries:

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

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

\int_{0}^{\frac{π}{3}} 2 cos (x) + 1 d x

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

\int_{0}^{4} g (x) d x

\int_{x^{2}}^{0} x cos (t^{2}) d t

\int_{0}^{9} 9 - x d x

What is the integral from \sqrt[3]{x} to pi of sin(t^3) ?
The integral from \sqrt[3]{x} to pi of sin(t^3) is