Evaluate the integral? #int e^(2x) cosx dx #
3 Answers
See below.
Explanation:
Using the de Moivre's identity
Taking the real part of this integral we have
I have solved with another way:
# int \ e^(2x) \cosx \ dx = 2/5e^(2x)cosx + 1/5e^(2x)sinx + C #
Explanation:
Let:
# I = int \ e^(2x) \cosx \ dx #
We can use integration by parts:
Let
# { (u,=cosx, => (du)/dx,=-sinx), ((dv)/dx,=e^(2x), => v,=1/2e^(2x) ) :}#
Then plugging into the IBP formula:
# int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx #
gives us
# int \ (cosx)(e^(2x)) \ dx = (cosx)(1/2e^(2x)) - int \ (1/2e^(2x))(-sinx) \ dx #
# :. I = 1/2e^(2x)cosx + 1/2int \ e^(2x) \ sinx \ dx # .... [A]
At first it appears as if we have made no progress, as now the second integral is similar to
Let
# { (u,=sinx, => (du)/dx,=cosx), ((dv)/dx,=e^(2x), => v,=1/2e^(2x) ) :}#
Then plugging into the IBP formula, gives us:
# int \ (sinx)(e^(2x)) \ dx = (sinx)(1/2e^(2x)) - int \ (1/2e^(2x))(cosx) \ dx #
# :. int \ e^(2x) \ sinx \ dx = 1/2e^(2x)sinx - 1/2I #
Inserting this result into [A] we get:
# I = 1/2e^(2x)cosx + 1/2(1/2e^(2x)sinx - 1/2I) + A #
# :. I = 1/2e^(2x)cosx + 1/4e^(2x)sinx - 1/4I + A #
# :. 4I = 2e^(2x)cosx + e^(2x)sinx - I + 4A #
# :. 5I = 2e^(2x)cosx + e^(2x)sinx + 4A #
# :. I = 2/5e^(2x)cosx + 1/5e^(2x)sinx + C #