What is the integral of x^n sin(x) dx using integration by parts?

What is the integral of x^n sin(x) dx using integration by parts?

• A. -x^n cos(x) + n∫ x^(n-1) cos(x) dx + C
• B. sin(x^n) + C
• C. cos(x^n) + C
• D. x^n sin(x) + C

Answer: (A)

To find the integral of \( x^n \sin(x) \, dx \) using integration by parts, we rely on the integration by parts formula, which is given by: \[ \int u \, dv = uv - \int v \, du \] In this context, we choose: - \( u = x^n \) (hence \( du = n x^{n-1} \, dx \)) - \( dv = \sin(x) \, dx \) (which gives \( v = -\cos(x) \)) Now, applying the integration by parts formula results in: \[ \int x^n \sin(x) \, dx = -x^n \cos(x) - \int -\cos(x) n x^{n-1} \, dx \] This simplifies to: \[ \int x^n \sin(x) \, dx = -x^n \cos(x) + n \int x^{n-1} \cos(x) \, dx \] The first term, \( -x^n \cos(x) \), is the product of our initial choice for \( u \) and \( v \

To find the integral of ( x^n \sin(x) , dx ) using integration by parts, we rely on the integration by parts formula, which is given by:

[

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

]

In this context, we choose:

• ( u = x^n ) (hence ( du = n x^{n-1} , dx ))

• ( dv = \sin(x) , dx ) (which gives ( v = -\cos(x) ))

Now, applying the integration by parts formula results in:

[

\int x^n \sin(x) , dx = -x^n \cos(x) - \int -\cos(x) n x^{n-1} , dx

]

This simplifies to:

[

\int x^n \sin(x) , dx = -x^n \cos(x) + n \int x^{n-1} \cos(x) , dx

]

The first term, ( -x^n \cos(x) ), is the product of our initial choice for ( u ) and ( v \