1. ## 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?

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

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

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

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

Find $\displaystyle du$ and $\displaystyle v$ and use the formula: $\displaystyle \int u~dv = uv - \int v~du$
this is the one derived from the product rule for differentiation right?

4. 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)