Compute ∫ (1/x) * sin(x) dx using parts.

Compute ∫ (1/x) * sin(x) dx using parts.

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

Answer: (A)

To compute the integral of \( \frac{1}{x} \sin(x) \, dx \) using integration by parts, we start with the formula for integration by parts: \[ \int u \, dv = uv - \int v \, du \] In this case, a suitable choice would be to let: - \( u = \sin(x) \) which gives \( du = \cos(x) \, dx \) - \( dv = \frac{1}{x} \, dx \) which leads to \( v = \ln|x| \) Applying integration by parts, we get: \[ \int \frac{1}{x} \sin(x) \, dx = \sin(x) \ln|x| - \int \ln|x| \cos(x) \, dx \] However, the correct simpler setup is to reidentify \( u \) and \( dv \): Let: - \( u = \frac{1}{x} \) which gives \( du = -\frac{1}{x^2} \, dx \) - \( dv = \sin(x) \, dx \) which results in \( v = -

To compute the integral of ( \frac{1}{x} \sin(x) , dx ) using integration by parts, we start with the formula for integration by parts:

[

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

]

In this case, a suitable choice would be to let:

• ( u = \sin(x) ) which gives ( du = \cos(x) , dx )

• ( dv = \frac{1}{x} , dx ) which leads to ( v = \ln|x| )

Applying integration by parts, we get:

[

\int \frac{1}{x} \sin(x) , dx = \sin(x) \ln|x| - \int \ln|x| \cos(x) , dx

]

However, the correct simpler setup is to reidentify ( u ) and ( dv ):

Let:

• ( u = \frac{1}{x} ) which gives ( du = -\frac{1}{x^2} , dx )

• ( dv = \sin(x) , dx ) which results in ( v = -