Question

Use integration by parts to solve the integral of (e^x)(Cos(2x))

Answer 1

\[\int\limits_{}^{} e ^{x}.\cos 2x\]

Answer 2

Do you know about Integration by parts??

Answer 3

\[\int\limits uv = u \int\limits v - \int\limits (u' \times \int\limits v)\]

Answer 4

yes I know the formula but both functions e^x and cos2x are both decaying functions when you derive or intergrate them. My guess is you'd have to use trig identities on cos2x

Answer 5

INT udv = uv - INT vdu

where u and v are functions

Answer 6

See here u = cos(2x) and v = \(e^x\)

Answer 7

yes

Answer 8

Can you use it in the formula???

Answer 9

Firstly let:

\[\int\limits e^x. \cos2x .dx = I\]

Answer 10

Now apply the formula..

Answer 11

yes you end up with. \[e ^{x}.-2\sin (2x) - \int\limits \left[ -2\sin (2x) . e ^{x} \right]\]

Answer 12

where you still have an integral function..

Answer 13

Where from - come from??

Answer 14

sorry. I did my integral wrong

Answer 15

Firstly tell me what is the integral of cos2x??

Answer 16

(sin2x)/2

Answer 17

Yes now write it properly in your notebook..

Answer 18

yes and you're still left with an integral function.

Answer 19

\[I = e ^{x}.\frac{\sin (2x)}{2} - \int\limits\limits\limits \left[e^x. \frac{\sin (2x)}{2}dx\right]\]

Answer 20

In the case we have one exponential and one trigonometric function in the Integral we have to do integration twice.

One time you have done this..

Now apply the integration to the last term again..

Answer 21

take the 2 out of the integral in last term..

Answer 22

do you mean take out 1/2?

Answer 23

Solve for the last term:

\[\int\limits e^x.\sin2xdx\]

u = sin2x and v = \(e^x\)

Again take the integral..

Doing or not??

Answer 24

Yes I mean that take 1/2 out of the integral..

Answer 25

okay um give me a sec

Answer 26

Take your time..

Answer 27

is it \[\sin 2x.e ^{x} - \int\limits \cos2x-e ^{x}\]

Answer 28

where you derive sin2x and integrate e^x

Answer 29

Can you find your mistakes??

Answer 30

integral of sin2x is : \(\large \frac{-cos2x}{2}\)

Answer 31

\[\sin2x.e ^{x} - \int\limits \cos2x.e ^{x}\]

Answer 32

oh, i derived sin2x not integrated it.

Answer 33

In first term you have to take integral of sin2x as you did in second..

Answer 34

i think you meant to say v = sin2x and u = ex instead of the other way around

Answer 35

No..

Answer 36

i'm confused, in the second one was i meant to derive or integrate sin2x?

Answer 37

SOrry..

Answer 38

Wait..

Answer 39

ok

Answer 40

We are doing it wrong..

Answer 41

Let me show the steps just wait..

Answer 42

okay