What is the integral from x to 5 of sin(t^3) ?

The integral from x to 5 of sin(t^3) is

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integral from x to 5 of sin(t^3)

\int_{x}^{5} sin (t^{3}) d t

- \frac{1}{2} i (\frac{5Γ ( \frac{1}{3} , i 5 ^{3} )}{3 3 i 5 ^{3}} - \frac{5Γ ( \frac{1}{3} , - i 5 ^{3} )}{3 3 - i 5 ^{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}}))

Solution steps

\int_{x}^{5} 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}})]_{x}^{5}

Compute the boundaries:

- \frac{1}{2} i (\frac{5Γ ( \frac{1}{3} , i 5 ^{3} )}{3 3 i 5 ^{3}} - \frac{5Γ ( \frac{1}{3} , - i 5 ^{3} )}{3 3 - i 5 ^{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}}))

= - \frac{1}{2} i (\frac{5Γ ( \frac{1}{3} , i 5 ^{3} )}{3 3 i 5 ^{3}} - \frac{5Γ ( \frac{1}{3} , - i 5 ^{3} )}{3 3 - i 5 ^{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}}))

\int tan^{4} (9 x) d x

d er i v a t i v e f (x) = \frac{2}{x} - \frac{1}{x ^{2}}

\int (3 x + \frac{1}{x}) d x

t an g e n t f (x) = \frac{x}{1 + x ^{2}}, (2, 0.4)

\frac{d}{d x} ((1 - sin (x))^{- \frac{1}{2}})

What is the integral from x to 5 of sin(t^3) ?
The integral from x to 5 of sin(t^3) is