A mathematical question for all you mathematicians

Hi.
I'll get straight to the point:

How do you integrate : (cos x)(e^x) ???

I just can't do it. Don't think it's possible with only P3 methods.
Could anyone outline each step to solving this problem, please? :smile:

Reply 1

visesh

17

dokebi

Hi.
I'll get straight to the point:

How do you integrate : (cos x)(e^x) ???

I just can't do it. Don't think it's possible with only P3 methods.
Could anyone outline each step to solving this problem, please? :smile:

by parts i think. diff e^x, int cos x

Reply 2

Gaz031

15

dokebi

Hi.
I'll get straight to the point:

How do you integrate : (cos x)(e^x) ???

I just can't do it. Don't think it's possible with only P3 methods.
Could anyone outline each step to solving this problem, please? :smile:

Using IPB: INT Vdu/dx dx = uv - INT udv/dx dx
v=e^x,dv/dx = e^x, du/dx = cosx, u=sinx.
INT (cosx)(e^x) dx = (e^x)(sinx) - INT (sinx)(e^x) dx
Using IBP again on the integral on the far right.
v=e^x, dv/dx=e^x. du/dx=sinx, u=-cosx.
INT (cosx)(e^x) dx = (e^x)(sinx) - [(-cosx)(e^x) - INT (-cosx)(e^x) dx ]
INT (cosx)(e^x) dx = (e^x)(sinx) + (cosx)(e^x) - INT (cosx)(e^x) dx
2 INT (cosx)(e^x) dx = (e^x)[sinx+cosx]
INT(cosx)(e^x) dx = (1/2)(e^x)(sinx+cosx).

By the way: As the trig ratio is cyclic and the exp function stays the same it doesn't matter which you differentiate and which you integrate - as long as you chose the same type of function for each one in each use of IBP (otherwise you'll just return to what you had at the start).

Reply 3

dokebi

OP

oh~~~!
thanks...
so u have to use integration by parts twice... and also remember to follow the non-cyclic pathway... didnt think of that...
i just kept getting lots of trigonomial functions and thought what the heck..

THANKS VERY MUCH!

Reply 4

username9816

14

dokebi

Find Int. (cos x)(e^x) dx

Let u = e^x --> du/dx = e^x
Let dv/dx = cosx --> v = sinx

Hence: Int. (cos x)(e^x) dx
= e^x.sinx - Int. sinx.e^x dx.

To find Int. sinx.e^x dx:
Let u = e^x --> du/dx = e^x
Let dv/dx = sinx --> v = -cosx

Hence: Int. sinx.e^x dx.
= -cosx.e^x + Int. e^x cosx.

Hence: Int. (cos x)(e^x) dx
= e^x.sinx - [-cosx.e^x + Int. e^x cosx]
= e^x.sinx + cosx.e^x - Int. e^x cosx

--> 2Int. cosx.e^x dx = e^x.sinx + e^x.cosx = e^x(sinx + cosx)
--> Int. cosx.e^x dx = [e^x(sinx + cosx)]/2

Reply 5

dokebi

OP

Hi!
I've got another question for tsr mathematicians....

I got stuck doing some Differential Equations.. (Mechanics 4)
It's basically about water flow, where I've got to apply Torricelli's Law to find an equation that can predict the height of water inside a container, where water is flowing in at a certain rate and flowing out at a certain rate simultaneously.
The equation I have to solve is this:

y=height of water, t=time, K and L = a constant

dy/dt + Ly^(1/2) = K/2{-Kt + 40^(1/2)}

I have to get an equation as: y= ......

The methods at my disposal (only 1/3 way thru Mechanics 4) cant solve this... at least I don't think so...
I'm thinking of using a numerical method... Euler's method... but ideally an algebraic method would be preferred...

Answers please!!! :biggrin:

A mathematical question for all you mathematicians

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