How do you integrate #int xsinx# by integration by parts method?

Question

Cameron Alexander · Accepted Answer

# int xsinx dx = sinx -xcosx + c #

If you are studying maths, then you should learn the formula for Integration By Parts (IBP), and practice how to use it:

# intu(dv)/dxdx = uv - intv(du)/dxdx #, or less formally # intudv=uv-intvdu #

I was taught to remember the less formal rule in word; "The integral of udv equals uv minus the integral of vdu". If you struggle to remember the rule, then it may help to see that it comes a s a direct consequence of integrating the Product Rule for differentiation.

Essentially we would like to identify one function that simplifies when differentiated, and identify one that simplifies when integrated (or is at least is integrable).

So for the integrand #xsinx#, hopefully you can see that #x# simplifies when differentiated and #sinx# effectively remains unchanged (#cosx# is still a trig function) under differentiation or integration.

Let # { (u=x, => , (du)/dx=1), ((dv)/dx=sinx, =>, v=-cosx ) :}#

Then plugging into the IBP formula gives us: # int(u)((dv)/dx)dx = (u)(v) - int(v)((du)/dx)dx # # :. int xsinx dx = (x)(-cosx) - int(-cosx)(1)dx # # " " = -xcosx + intcosxdx # # " " = -xcosx + sinx + c #

Samantha Adkison · Answer

undefined To integrate $ \int x \sin(x) $ using integration by parts, follow these steps:

1. Choose $ u $ and $ dv $:
   Let $ u = x $ and $ dv = \sin(x) \, dx $.

2. Calculate $ du $ and $ v $:
   Differentiate $ u $ to get $ du $, which is $ du = dx $, and integrate $ dv $ to get $ v $, which is $ v = -\cos(x) $.

3. Apply the integration by parts formula:
   $$ \int u \, dv = uv - \int v \, du $$

4. Substitute the values:
   $$ \int x \sin(x) \, dx = -x \cos(x) - \int (-\cos(x)) \, dx $$

5. Simplify and integrate:
   $$ \int x \sin(x) \, dx = -x \cos(x) + \int \cos(x) \, dx $$

6. Integrate $ \cos(x) $ to get the final result:
   $$ \int x \sin(x) \, dx = -x \cos(x) + \sin(x) + C $$
   where $ C $ is the constant of integration.