Question

Evaluate the indefinite integral of e^(4x)sin(2x) dx

Answer 1

integration by parts.. ughh longgggg

Answer 2

http://www.wolframalpha.com/input/?i=integrate+%28e^%284x%29sin%282x%29+dx%29
click on show steps

Answer 3

We have
\[\int e^{4x} \sin 2x dx\]
we'll use integration by parts
\[\int uv=u\int v- \int (v' \int u)\]
here
\[I=\int e^{4x} \sin 2x dx\]

here
\[v=e^4x\]
\[u=\sin2 x dx\]
we now integrate I using by parts
\[I= e^{4x}\int\sin 2x dx- \int 4e^{4x }(\int \sin 2x dx) dx\]
we have
\[ I = -e^{4x} \cos 2x /2+ \int 4 e^{4x} \cos 2x /2dx\]
we have now
\[I=-e^{4x} \cos 2x /2+ \int 2e^{4x} \cos 2x dx\]
let's integrate the second term using by parts
\[I=-e^{4x} \cos 2x /2+ 2e^{4x}\int \cos 2x dx-\int \frac{d}{dx}2e^{4x} (\int \cos 2x dx) dx\]
we get now
\[I=-e^{4x} \cos 2x /2+ 2e^{4x}\sin 2x /2-\int 8 e^{4x} \sin 2x /2 dx\]
we know that \[I=\int e^{4x} \sin 2x dx\]
the second term is -4I
so we get
\[I=-e^{4x} \cos 2x /2+ 2e^{4x}\sin 2x /2-4I\]
so we get
\[5I=-e^{4x} \cos 2x /2+ 2e^{4x}\sin 2x /2\]
we get finally
\[I=\frac{e^{4x}}{5}(-\cos 2x /2+ \sin 2x )\]

Answer 4

forgot to add constant , add C with the answer