What is the integral from 0 to t^2 of cos(t^3) ?

The integral from 0 to t^2 of cos(t^3) is

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integral from 0 to t^2 of cos(t^3)

\int_{0}^{t^{2}} cos (t^{3}) d t

- \frac{t ^{2} ( 3 i ( t ^{2} ) ^{3} Γ ( \frac{1}{3} , - i ( t ^{2} ) ^{3} ) + 3 - i ( t ^{2} ) ^{3} Γ ( \frac{1}{3} , i ( t ^{2} ) ^{3} ) )}{6 3 ( t ^{2} ) ^{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}^{t^{2}} 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}}]_{0}^{t^{2}}

Compute the boundaries:

- \frac{t ^{2} ( 3 i ( t ^{2} ) ^{3} Γ ( \frac{1}{3} , - i ( t ^{2} ) ^{3} ) + 3 - i ( t ^{2} ) ^{3} Γ ( \frac{1}{3} , i ( t ^{2} ) ^{3} ) )}{6 3 ( t ^{2} ) ^{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{t ^{2} ( 3 i ( t ^{2} ) ^{3} Γ ( \frac{1}{3} , - i ( t ^{2} ) ^{3} ) + 3 - i ( t ^{2} ) ^{3} Γ ( \frac{1}{3} , i ( t ^{2} ) ^{3} ) )}{6 3 ( t ^{2} ) ^{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}}

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What is the integral from 0 to t^2 of cos(t^3) ?
The integral from 0 to t^2 of cos(t^3) is