What is the result of integrating xsin(x) dx using integration by parts?

What is the result of integrating xsin(x) dx using integration by parts?

• A. -xcos(x) + ∫ cos(x) dx + C
• B. xe^x + C
• C. cos(x) + C
• D. -sin(x) + C

Answer: (A)

To determine the result of integrating \( x \sin(x) \, dx \) using integration by parts, we apply the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] In this case, we can choose \( u = x \) and \( dv = \sin(x) \, dx \). Then, we compute \( du \) and \( v \): - The derivative \( du = dx \) - The integral \( v = -\cos(x) \) Now, substituting these into the integration by parts formula: \[ \int x \sin(x) \, dx = x (-\cos(x)) - \int (-\cos(x)) \, dx \] This simplifies to: \[ -x \cos(x) + \int \cos(x) \, dx \] Next, we integrate \( \cos(x) \): \[ \int \cos(x) \, dx = \sin(x) \] So, putting it all together, we have: \[ \int x \sin(x) \, dx = -x \cos(x) + \sin(x) + C \] Since the

To determine the result of integrating ( x \sin(x) , dx ) using integration by parts, we apply the integration by parts formula:

[

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

]

In this case, we can choose ( u = x ) and ( dv = \sin(x) , dx ). Then, we compute ( du ) and ( v ):

• The derivative ( du = dx )

• The integral ( v = -\cos(x) )

Now, substituting these into the integration by parts formula:

[

\int x \sin(x) , dx = x (-\cos(x)) - \int (-\cos(x)) , dx

]

This simplifies to:

[

-x \cos(x) + \int \cos(x) , dx

]

Next, we integrate ( \cos(x) ):

[

\int \cos(x) , dx = \sin(x)

]

So, putting it all together, we have:

[

\int x \sin(x) , dx = -x \cos(x) + \sin(x) + C

]

Since the