How do you integrate #int xcos(3x)# by integration by parts method?

Question

William Abdalla · Accepted Answer

The goal of integration by parts is to differentiate one of the terms in the product so that it becomes #1#, leaving only one term to be integrated. Take a look at this problem to understand better.

Let #u = x# and#dv = cos(3x)dx#. To apply the integration by parts formula, we need #v# and #du#. #du = dx#. We can find #v# through #u#-substitution.

Let #u_2 = 3x#. Then #du_2 = 3dx# and #dx = (du_2)/3#

We can rewrite:

#=int(cosu_2) * (du_2)/3#

#=1/3intcosu_2du_2#

#=1/3sinu_2#

Since #u_2 = 3x#:

#=1/3sin(3x)#

Thus, #v = 1/3sin(3x)#.

The integration by parts formula is #int(udv) = uv - int(vdu)#.

#int(xcos(3x)) dx= 1/3sin(3x) * x - int(1 * 1/3sin(3x)dx)#

#int(xcos3x)dx = 1/3xsin(3x) - int(1/3sin(3x)dx)#

We repeat the substitution process performed above to integrate #1/3sin(3x)#

Let #u = 3x#. Then #du = 3dx# and #dx = 1/3du#

#=int1/3sinu * 1/3du#

#=1/9intsinu du#

#=-1/9cosu#

#=-1/9cos(3x)#

The integral of the entire expression is therefore:

#int(xcos3x)dx = 1/3xsin3x+ 1/9cos3x+ C#, where#C# is a constant.

Hopefully this helps!

Isaiah Allen · Answer

undefined To integrate ∫x*cos(3x) using integration by parts, you assign one part of the function as u and the other part as dv/dx. The formula for integration by parts is ∫u dv = uv - ∫v du.

Assign u = x and dv = cos(3x)dx. Then, differentiate u to find du/dx, and integrate dv to find v.

Next, apply the integration by parts formula, substituting the values of u, v, du/dx, and dv into the formula.

Finally, solve the resulting integral.