# integration problem

• Nov 8th 2009, 09:20 PM
yen yen
integration problem
this is a L-R Series Circuit with
V= 240 cos 10t
R= 100 Ohm
L= 20 Henry

It asks to find current in circuit at any time if initially there is no current.

this is from my lecture notes so there is a step im not sure how to get.
d/dt(ie^5t) = 12 e^5t cos 10t
ie^5t = 12 ∫e^5t cos 10t dt
= 12/25 e^5t [cos 10t + 2sin 10t] +c

how do i integrate e^5t cos 10t to get to the next step?
• Nov 8th 2009, 09:43 PM
Jhevon
Quote:

Originally Posted by yen yen
this is a L-R Series Circuit with
V= 240 cos 10t
R= 100 Ohm
L= 20 Henry

It asks to find current in circuit at any time if initially there is no current.

this is from my lecture notes so there is a step im not sure how to get.
d/dt(ie^5t) = 12 e^5t cos 10t
ie^5t = 12 ∫e^5t cos 10t dt
= 12/25 e^5t [cos 10t + 2sin 10t] +c

how do i integrate e^5t cos 10t to get to the next step?

you can use integration by parts. do you remember the formula?

Let $u=\cos 10t$ and $dv = e^{5t}~dt$

Find $du$ and $v$ and use the formula: $\int u~dv = uv - \int v~du$
• Nov 8th 2009, 09:49 PM
yen yen
Quote:

Originally Posted by Jhevon
you can use integration by parts. do you remember the formula?

Let $u=\cos 10t$ and $dv = e^{5t}~dt$

Find $du$ and $v$ and use the formula: $\int u~dv = uv - \int v~du$

this is the one derived from the product rule for differentiation right?
• Nov 8th 2009, 09:52 PM
Jhevon
Quote:

Originally Posted by yen yen
this is the one derived from the product rule for differentiation right?

that's correct. we use it to integrate products (where possible)