Question

x cos(4x)dx
u = x
dv = cos 4xdx
@LeibyStrauss

Answer 1

use integration by parts
u = dx ---> du = 1
dv = -cos4xdx --- > v = - (1/4) sin 4 x

INT udv = uv - INT vdu

Answer 2

\[\int\limits_{}^{}\cos(4x)dx\]

u = 4x
du = 4dx
1/4du = dx

\[\frac{ 1 }{ 4 }\int\limits_{}^{} \cos(u)dx = \frac{ 1 }{ 4 }\sin(4x)\]

\[v = \frac{ 1 }{ 4 }\sin(4x)\]

Answer 3

\[\int\limits_{}^{} u dv = uv- \int\limits_{}^{} v du\]

Answer 4

\[x \frac{ 1 }{ 4 } \sin(4x) - \int\limits \frac{ 1 }{ 4 } \sin(4x)\]

now let's solve

\[\int\limits \frac{ 1 }{ 4 } \sin(4x)\]

u = 4x
du = 4dx
1/4 du = dx

\[\frac{ 1 }{ 4 }\int\limits \sin(u)\]

Answer 5

\[\frac{ 1 }{ 4 }[-\cos(u)] = -\frac{ 1 }{ 4 }\cos(4x)\]

Answer 6

reverse of the product rule on the derivatives

[f*g] ' = f ' *g +g ' *f

parts is the integration of that

Answer 7

@vkarwee I made a mistake somewhere.

Answer 8

i think you forgot to multiply

Answer 9

I forgot to pull out the 1/4 and then multiply by 1/4 again. I'll redo it

Answer 10

\[1/4\int\limits_{}^{}\sin(4x)\]

Answer 11

sin(4x) / 16 + x* cos(4x)/4

Answer 12

= \[= -\frac{ 1 }{ 16 }\sin(4x)\]

Answer 13

sometimes, if you can have the answer, with a calculator or whatever first, it may help you figure out the method, from the answer form

Answer 14

Final answer

\[\frac{ x }{ 4 } \sin(4x) + \frac{ 1 }{ 16 }\cos(4x)\]

Answer 15

the sin and cos is mixed around,

Answer 16

@DanJS

v = 1/4 sin(4x)

\[\int\limits v du = \frac{ 1 }{ 16 } \cos(4x)\]

Answer 17

forgot a the minus

Answer 18

forget i was evfen at this question, i did the original prob

x*cos(4x) to get that answer, intresting how they are related though

Answer 19

**x*sin(4x) is what i started with

my mistake

Answer 20

if you need to know integral of x*sin(4x) , you know how it is related to the cos version