What is the integral from 0 to 1 of t(1+it^2)^3 ?

The integral from 0 to 1 of t(1+it^2)^3 is i 5/8

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integral from 0 to 1 of t(1+it^2)^3

\int_{0}^{1} t (1 + i t^{2})^{3} d t

i \frac{5}{8}

Solution steps

\int_{0}^{1} t (1 + i t^{2})^{3} d t

Expand

t (1 + i t^{2})^{3} : (t - 3 t^{5}) + i (- t^{7} + 3 t^{3})

= \int_{0}^{1} t - 3 t^{5} + i (- t^{7} + 3 t^{3}) d t

Apply the Sum Rule:

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

= \int_{0}^{1} t d t - \int_{0}^{1} 3 t^{5} d t + \int_{0}^{1} i (- t^{7} + 3 t^{3}) d t

\int_{0}^{1} t d t = \frac{1}{2}

\int_{0}^{1} 3 t^{5} d t = \frac{1}{2}

\int_{0}^{1} i (- t^{7} + 3 t^{3}) d t = i \frac{5}{8}

= \frac{1}{2} - \frac{1}{2} + i \frac{5}{8}

Simplify

\frac{1}{2} - \frac{1}{2} + i \frac{5}{8} : i \frac{5}{8}

= i \frac{5}{8}

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What is the integral from 0 to 1 of t(1+it^2)^3 ?
The integral from 0 to 1 of t(1+it^2)^3 is i 5/8