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

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

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

\int_{3 x}^{3 π} cos (t^{3}) d t

- \frac{3 π ( 3 i ( 3 π ) ^{3} Γ ( \frac{1}{3} , - i ( 3 π ) ^{3} ) + 3 - i ( 3 π ) ^{3} Γ ( \frac{1}{3} , i ( 3 π ) ^{3} ) )}{6 3 ( 3 π ) ^{6}} - - \frac{3 x ( 3 i ( 3 x ) ^{3} Γ ( \frac{1}{3} , - i ( 3 x ) ^{3} ) + 3 - i ( 3 x ) ^{3} Γ ( \frac{1}{3} , i ( 3 x ) ^{3} ) )}{6 3 ( 3 x ) ^{6}}

Solution steps

\int_{3 x}^{3 π} cos (t^{3}) d t

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{t ( 3 i t ^{3} Γ ( \frac{1}{3} , - i t ^{3} ) + 3 - i t ^{3} Γ ( \frac{1}{3} , i t ^{3} ) )}{6 3 t ^{6}}]_{3 x}^{3 π}

Compute the boundaries:

- \frac{3 π ( 3 i ( 3 π ) ^{3} Γ ( \frac{1}{3} , - i ( 3 π ) ^{3} ) + 3 - i ( 3 π ) ^{3} Γ ( \frac{1}{3} , i ( 3 π ) ^{3} ) )}{6 3 ( 3 π ) ^{6}} - - \frac{3 x ( 3 i ( 3 x ) ^{3} Γ ( \frac{1}{3} , - i ( 3 x ) ^{3} ) + 3 - i ( 3 x ) ^{3} Γ ( \frac{1}{3} , i ( 3 x ) ^{3} ) )}{6 3 ( 3 x ) ^{6}}

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

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