integral from 0 to x of sin(t^3+1)

解答

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

sin (1) - \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{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}} + cos (1) (- \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{0 \cdot Γ ( \frac{1}{3} , i 0 ^{3} )}{3 3 i 0 ^{3}} - \frac{0 \cdot Γ ( \frac{1}{3} , - i 0 ^{3} )}{3 3 - i 0 ^{3}})))

求解步骤

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

使用三角恒等式改写

= \int_{0}^{x} cos (t^{3}) sin (1) + cos (1) sin (t^{3}) d t

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{x} cos (t^{3}) sin (1) d t + \int_{0}^{x} cos (1) sin (t^{3}) d t

\int_{0}^{x} cos (t^{3}) sin (1) d t = sin (1) - \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{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}}

\int_{0}^{x} cos (1) sin (t^{3}) d t = cos (1) (- \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{0 \cdot Γ ( \frac{1}{3} , i 0 ^{3} )}{3 3 i 0 ^{3}} - \frac{0 \cdot Γ ( \frac{1}{3} , - i 0 ^{3} )}{3 3 - i 0 ^{3}})))

= sin (1) - \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{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}} + cos (1) (- \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{0 \cdot Γ ( \frac{1}{3} , i 0 ^{3} )}{3 3 i 0 ^{3}} - \frac{0 \cdot Γ ( \frac{1}{3} , - i 0 ^{3} )}{3 3 - i 0 ^{3}})))

\int_{1}^{2} (x + 3)^{5} d x

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