intergration

intergrate between o and pi/2 e^3x cos 4x dx

have no idea please help

thanks loads

You could integrate by parts twice. Or you could do it this way:

(d/dx) e^(3x) sin(4x)
= e^(3x)(3sin(4x) + 4cos(4x))

(d/dx) e^(3x) cos(4x)
= e^(3x)(-4sin(4x) + 3cos(4x))

--

(d/dx) e^(3x) (4sin(4x) + 3cos(4x))
= 25 e^(3x) cos(4x)

--

(int from 0 to pi/2) e^(3x) cos(4x) dx
= [(1/25)e^(3x) (4sin(4x) + 3cos(4x))] (from 0 to pi/2)
= (3/25)(e^(3pi/2) - 1)

Reply 2

username9816

sarah12345

intergrate between 0 and pi/2 e^3x cos 4x dx

Let u = cos4x -> du/dx = -4sin4x
Let dv/dx = e^3x -> v = (e^3x)/3

Hence: Int. e^3x.cos4x dx = (e^3x.cos4x)/3 + 4/3 Int. (sin4x.e^3x) dx.

To find Int. e^3x.sin4x dx.
Let u = sin4x -> du/dx = 4cos4x
Let dv/dx = e^3x -> v = (e^3x)/3

Hence: Int. e^3x.sin4x dx. = (e^3x.sin4x)/3 - 4/3 Int. e^3x.cos4x dx.

Therefore: Int. e^3x.cos4x dx = (e^3x.cos4x)/3 + 4/3{(e^3x.sin4x)/3 - 4/3 Int. e^3x.cos4x dx.} = (e^3x.cos4x)/3 + (4/9)(e^3x.sin4x) - (16/9)Int. e^3x.cos4x dx.

-> (16/9 + 1) Int. e^3x.cos4x dx. = (e^3x.cos4x)/3 + (4/9)(e^3x.sin4x)
-> (25/9) Int. e^3x.cos4x dx. = (3e^3x.cos4x)/9 + (4/9)(e^3x.sin4x)
-> Int. e^3x.cos4x dx. = (3e^3x.cos4x + 4e^3x.sin4x)/25
-> Int. e^3x.cos4x dx. = [e^3x(3cos4x + 4sin4x)]/25

Therefore: Int. (0 to Pi/2) e^3x.cos4x dx. = 1/25[e^3x(3cos4x + 4sin4x)] (Limits 0 and Pi/2) = 1/25{[3e^(3Pi/2)] - [3]} = (3/25)[e^(3Pi/2) - 1]

-> Int. (0 to Pi/2) e^3x.cos4x dx. = (3/25)[e^(3Pi/2) - 1]

Reply 3

k@tie

You need to choose different parts each time.
Call the integral I
Then, choosing
u=e^3x => u'=3e^3x
v'=cos4x => v=(1/4)sin4x

Then I = (1/4)[sin4x . e^3x] - integral (3/4)(sin4x . e^3x) *A*

Choosing different parts gives you
u=cos4x => u'=-4sin4x
v'=e^3x => v=(1/3)e^3x

Then I=(1/3)[cos 4x . e^3x] + integral (4/3) (sin4x . e^3x)

Multiply through this expression by 9/16 (you'll see why in a sec)
(9/16)I = (9/16)(1/3)[cos 4x . e^3x] + integral (4/3)(9/16)(sin4x . e^3x)
(9/16)I = (9/48)[cos 4x . e^3x] + integral (3/4)(sin4x . e^3x) *B*

You now have two different expressions for I, add the original one and the new one (with the 9/16 in) (which is *a* + *b*)
To give (25/16)I = (1/4)[sin4x . e^3x] - integral (3/4)(sin4x . e^3x) + (9/48)[cos 4x . e^3x] + integral (3/4)(sin4x . e^3x)
=> (25/16)I = (1/4)[sin4x . e^3x] + (9/48)[cos 4x . e^3x]
Then pop in your limits and divide by 25/16 to find the area I

thankyou soooo much

Jonny W

What is that method? I don't understand what you've done there!

Reply 6

dvs

We want to find ∫ f(x) dx; Jonny_W found, with some manipulation, a function g(x) such that:
dg/dx = kf(x), where k is some constant.

So, loosely:
f(x) = (1/k) dg/dx
∫ f(x) dx = (1/k) g(x) + C

Jonny's steps were:

Find two functions to differentiate, and use the fact that (cosx)'=-sinx so that adding their derivatives together would cancel some terms out:
e^(3x) sin(4x) & e^(3x) cos(4x)

Differentiate those functions:
(d/dx) e^(3x) sin(4x)
= e^(3x)(3sin(4x) + 4cos(4x))

(d/dx) e^(3x) cos(4x)
= e^(3x)(-4sin(4x) + 3cos(4x))

We want to lose the sin(4x) term. So multiply the first one by 4 and the second one by 3, and then add the functions together:
e^(3x) (4sin(4x) + 3cos(4x))
And add up the derivatives:
25 e^(3x) cos(4x)

So:
(d/dx) e^(3x) (4sin(4x) + 3cos(4x))
= 25 e^(3x) cos(4x)

In the original question, we wanted to find:
I = ∫ e^(3x) cos(4x) dx

Which equals:
I = (1/25) ∫ e^(3x) 25cos(4x) dx

We already know what function to differentiate to get e^(3x) 25cos(4x). So:
I = (1/25) . [e^(3x) (4sin(4x) + 3cos(4x))] + C

intergration

Related discussions

Latest

Trending

Trending

Articles for you