What is the integral of e^{i2t} ?

The integral of e^{i2t} is 1/2 sin(2t)-i 1/2 cos(2t)+C

Popular >

integral of e^{i2t}

\int e^{i 2 t} d t

\frac{1}{2} sin (2 t) - i \frac{1}{2} cos (2 t) + C

Solution steps

\int e^{i 2 t} d t

Apply imaginary number rule:

e^{ia} = cos (a) + i sin (a)

= \int cos (2 t) + i sin (2 t) 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 cos (2 t) d t + \int i sin (2 t) d t

\int cos (2 t) d t = \frac{1}{2} sin (2 t)

\int i sin (2 t) d t = - i \frac{1}{2} cos (2 t)

= \frac{1}{2} sin (2 t) - i \frac{1}{2} cos (2 t)

Add a constant to the solution

= \frac{1}{2} sin (2 t) - i \frac{1}{2} cos (2 t) + C

\int tan^{4} (2 x) sec (2 x) d x

\int \frac{6 x ^{2} + x + 16}{( x ^{2} + 3 ) ( x - 2 )} d x

\int (\frac{6 + x ^{2}}{x}) d x

\int tan (x) + sec^{2} (x) d x

\int \frac{e 4 x + 2}{4 x + 2} d x

What is the integral of e^{i2t} ?
The integral of e^{i2t} is 1/2 sin(2t)-i 1/2 cos(2t)+C