What is the integral of e^{it} ?

The integral of e^{it} is sin(t)-icos(t)+C

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integral of e^{it}

\int e^{i t} d t

sin (t) - i cos (t) + C

Solution steps

\int e^{i t} d t

Apply u-substitution

: \int - i e^{u} d u

= \int - i e^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - i \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - i e^{u}

Substitute back

u = i t

= - i e^{i t}

Simplify

- i e^{i t} : sin (t) - i cos (t)

= sin (t) - i cos (t)

Add a constant to the solution

= sin (t) - i cos (t) + C

x \to \infty lim (\frac{7}{x} - 1)

y^{^{''}} + 5 y^{^{'}} + 4 y = sin (t)

\frac{d}{d x} ((\frac{x}{1 + x})^{5})

- 3 t (2 t^{4} - 1)^{7}

f (x) = ln (8 - 9 x)