How do you find the integral of #intxcos(5x)dx#?

2 Answers
Mar 23, 2017

Use integration by parts.
#intudv=uv-intvdu#

Explanation:

Let #u = x#, then #du = dx, dv=cos(5x)dx, and v=1/5sin(5x)#

Substituting these values into the integration by parts equation:

#intxcos(5x)dx=x/5sin(5x)-1/5intsin(5x)dx#

The remaining integral on the right is trivial:

#intxcos(5x)dx=x/5sin(5x)+1/25cos(5x)+C#

Mar 23, 2017

#1/25(5x sin 5x + cos 5x) + C#

Explanation:

Use integration by parts #int u dv = u*v - int v du#

Let #u = x, du = dx#
Let #dv = cos 5x dx, v = int cos 5x dx#

Let #w = 5x, dw = 5 dx#
#int cos 5x dx = 1/5 int cos w dw = 1/5 sin 5x#

Fill in the integration by parts:
#int x cos 5x dx = 1/5x sin 5x - 1/5 int sin 5xdx#

To complete the last integration piece:
Let #w = 5x, dw = 5 dx#

#-1/5int sin 5xdx = -1/5 *1/5 int 5sin 5x dx = -1/25 int sin w dw = -1/25 (-cos 5x) = 1/25 cos 5x#

The final solution simplified:
#int x cos 5x dx = 1/5x sin 5x +1/25 cos 5x + C= 1/25(5x sin 5x + cos 5x) + C#