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The integral intcos(log(e)x)dx is equal ...

The integral `intcos(log_(e)x)dx` is equal to: (where C is a constant of integration)

A

`x/2[sin(log_ex-cos(log_ex)]+C`

B

`x/2[cos(log_ex+sin(log_ex)]+C`

C

`x[cos(log_ex+sin(log_ex)]+C`

D

`x[cos(log_ex-sin(log_ex)]+C`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the integral \( \int \cos(\log x) \, dx \), we will use integration by parts. Let's go through the solution step by step. ### Step-by-Step Solution: 1. **Identify the integral**: \[ I = \int \cos(\log x) \, dx \] 2. **Use integration by parts**: We will use the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] Let's choose: - \( u = \cos(\log x) \) - \( dv = dx \) 3. **Differentiate and integrate**: Now we need to find \( du \) and \( v \): - Differentiate \( u \): \[ du = -\sin(\log x) \cdot \frac{1}{x} \, dx \] - Integrate \( dv \): \[ v = x \] 4. **Apply the integration by parts formula**: Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \[ I = x \cos(\log x) - \int x \left(-\sin(\log x) \cdot \frac{1}{x}\right) \, dx \] Simplifying gives: \[ I = x \cos(\log x) + \int \sin(\log x) \, dx \] 5. **Evaluate the new integral**: Let’s denote the new integral as \( J = \int \sin(\log x) \, dx \). We will again use integration by parts: - Choose: - \( u = \sin(\log x) \) - \( dv = dx \) - Then: - \( du = \cos(\log x) \cdot \frac{1}{x} \, dx \) - \( v = x \) 6. **Apply integration by parts again**: \[ J = x \sin(\log x) - \int x \cos(\log x) \cdot \frac{1}{x} \, dx \] Simplifying gives: \[ J = x \sin(\log x) - \int \cos(\log x) \, dx \] Notice that \( \int \cos(\log x) \, dx = I \), so we can substitute: \[ J = x \sin(\log x) - I \] 7. **Combine the results**: Now we substitute \( J \) back into the equation for \( I \): \[ I = x \cos(\log x) + (x \sin(\log x) - I) \] Rearranging gives: \[ 2I = x \cos(\log x) + x \sin(\log x) \] Therefore: \[ I = \frac{1}{2} (x \cos(\log x) + x \sin(\log x)) \] 8. **Final result**: The integral is: \[ \int \cos(\log x) \, dx = \frac{x}{2} (\cos(\log x) + \sin(\log x)) + C \] where \( C \) is the constant of integration.
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