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

Question

Paisley Johnson · Accepted Answer

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

Evelyn Agard · Answer

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

Ezra Adams · Answer

undefined To find the integral of $ \int x \cos(5x) \, dx $, you can use integration by parts. Apply the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

Let:
$$ u = x $$
$$ dv = \cos(5x) \, dx $$

Then, differentiate $ u $ to get $ du $, and integrate $ dv $ to get $ v $.

$$ du = dx $$
$$ v = \frac{1}{5} \sin(5x) $$

Now, apply the integration by parts formula:

$$ \int x \cos(5x) \, dx = x \cdot \frac{1}{5} \sin(5x) - \int \frac{1}{5} \sin(5x) \, dx $$

This integral can be straightforwardly integrated to get the final result.