Question

integral of xcos4x with explanation please?

Answer 1

use substitution

Answer 2

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

Answer 3

integration by parts

Answer 4

Is there any calculus problem that myin can't solve?

Answer 5

are u sure it's not 4x^2

Answer 6

myin, it's an indefinite integral

Answer 7

\[1/4*\sin 4x *x +1/16*\cos 4x \]

Answer 8

i know it is an indefinite integral
integration by parts is the way to do this one
the way i showed above

Answer 9

*thunder*

Answer 10

\[\int\limits_{}^{}x \sin 4xdx\]\[u = x, du = dx\]\[dv = sen(4x)dx\]\[v = (1/4)\cos (4x)\]\[\int\limits_{}^{}udv = uv - \int\limits_{}^{}vdu\]\[\int\limits_{}^{}x \sin 4xdx = x(1/4)\cos 4x - \int\limits_{}^{}(1/4)\cos(4x)dx\]= x(1/4)cos 4x - (1/4)(-(1/4)sen(4x))
=x(1/4)cos 4x + (1/16)sen(4x)