Evaluate the integral? #int e^(2x) cosx dx #

3 Answers
Jul 5, 2017

See below.

Explanation:

Using the de Moivre's identity

#e^(ix)=cosx+isinx# we have

#int e^(2x)cosx dx+i int e^(2x)sinx dx = int e^((2+i)x) dx = 1/(2+i)e^((2+i)x)+C=#

#=1/(2+i) e^(2x)(cosx+isin x)+C=#

#(2-i)/(2^2+1)(cos x+isinx) +C = #

#1/5(2cosx+sinx+i(2sinx-cosx))e^(2x)+C#

Taking the real part of this integral we have

#int e^(2x)cosx dx = 1/5(2cosx+sinx)e^(2x) + C#

I have solved with another way:
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Jul 5, 2017

# 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 #I#, having exchanged #cosx# for #sinx#, but if we apply IBP a second time then the progress will become clear:

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 #