Question

different practice problem...
if you want to know where they are from,
citing sources: https://math.dartmouth.edu/archive/m11f97/public_html/m12integrals.pdf

Answer 1

\[\int\limits_{ }^{ } e^x \cos(2x) dx\]
just posting

Answer 2

there is a standard formula for
\(\int e^{ax}\cos (bx) dx\)
but you can derive it by using uv rule twice :)

Answer 3

I clikced it again

Answer 4

I know how to do it I think

Answer 5

\[\int\limits_{ }^{ } e^x \cos(2x) dx=e^x \sin(2x) dx+2\int\limits_{ }^{ } e^x \sin(2x) dx\]I am differentiating the cos(2x).
\[\int\limits_{ }^{ } e^x \cos(2x) dx=e^x \sin(2x) dx+2e^x \sin(2x) -4\int\limits_{ }^{ } e^{x}\cos(2x) dx\]

Answer 6

please, try this another method:
\[\int\limits e ^{x}\cos(2x) dx=\int\limits e ^{x}\frac{ e ^{i2x} -e ^{-i2x}}{ 2 }dx\]
then consider the real part only of your result

Answer 7

\[\int\limits_{ }^{ } e^x \cos(2x) dx=\frac{e^x \sin(2x) dx+2e^x \sin(2x)}{5}\]

Answer 8

ooops \[\frac{ e ^{i2x}+e ^{-i2x} }{ 2 }\]

Answer 9

right?

Answer 10

you mena
e^x cos 2x in the first term of the numerator ?

Answer 11

*mean

Answer 12

\(\int\limits_{ }^{ } e^x \cos(2x) dx=\frac{e^x \cos(2x) dx+2e^x \sin(2x)}{5}\)

Answer 13

yes, I see where I made a mistake

Answer 14

yes, right, tnx for pointing it out.

Answer 15

try that other approach too :)

Answer 16

I wonder how Michelle got this

Answer 17

\(e^{ix} = \cos x + i \sin x\)

Answer 18

oh that...

Answer 19

but why would I ever need that?

Answer 20

I got this:
\[\frac{ e ^{x} }{ 5 }[\cos (2x)+2 \sin (2x)]\]
namely the same result of @hartnn
furthermore there is no imaginary part

Answer 21

yes same exact thing

Answer 22

when uv rule fails, and you have sin or cos in the integrand,
that will help you...

Answer 23

I have always delt with problems that are solvable by normal integration bp, substitution and at times partial fractions. I guess I will learn this in calc 2....

Answer 24

anyways, tnx again!

Answer 25

welcome ^_^