∫xcos(2x)dx
What I did wrong
u=x
du=dx
dv=cos(2x)dx
v=-sin(2x)/2
-xsin(2x)/2 +∫sin(2x)/2dx--> -xsin(2x)/2 -cos(2x)/4+C
oooup, this one why my answer wrong need steps and explain.
u=x
du=dx
dv=cos(2x)dx
v=sin(2x)/2
xsin(2x)/2 -∫sin(2x)/2dx--> = xsin(2x)/2 -cos(2x)/4+C
-∫sin(2x)/2dx--> = +cos(2x)/4
It seems like you made a mistake while integrating the term ∫sin(2x)/2 dx in your solution. Let's go over the correct steps to find the integral of ∫xcos(2x) dx using integration by parts.
Integration by parts formula:
∫u dv = uv - ∫v du
Let's choose u = x and dv = cos(2x) dx. This gives us:
du = dx (since the derivative of x with respect to x is 1)
v = ∫cos(2x) dx = (1/2)sin(2x) (using the integral of cos(x) which is sin(x))
Now we can apply the formula:
∫xcos(2x) dx = uv - ∫v du
= x * (1/2)sin(2x) - ∫(1/2)sin(2x) dx
Simplifying the integral:
∫(1/2)sin(2x) dx = (1/2) ∫sin(2x) dx
= (1/2) * (-1/2) * cos(2x) [Using the integral of sin(x) which is -cos(x)]
= -1/4 cos(2x)
Now we can substitute this result back into the integration by parts formula:
∫xcos(2x) dx = x * (1/2)sin(2x) - ∫(1/2)sin(2x) dx
= x * (1/2)sin(2x) + 1/4 cos(2x) + C
Therefore, the correct solution should be:
∫xcos(2x) dx = x * (1/2)sin(2x) + 1/4 cos(2x) + C, where C is the constant of integration.