Question

The value of \[ \int \sin(\log x) \, dx + \int \cos(\log x) \, dx \] is equal to
 

• A. \( \sin(\log x) + C \)
• B. \( \cos(\log x) + C \)
• C. \( x \sin(\log x) + C \)
• D. \( x \cos(\log x) + C \)

Hint:

Use integration by parts to solve integrals of the form \( \int \sin(\log x) \, dx \) and \( \int \cos(\log x) \, dx \).

Answer 1

The Correct Option is C

Step 1: Applying integration by parts.
To solve the integral \( \int \sin(\log x) \, dx \), we use integration by parts. Let: \[ u = \sin(\log x), \quad dv = dx \] Then: \[ du = \cos(\log x) \cdot \frac{1}{x} \, dx, \quad v = x \] Applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get: \[ \int \sin(\log x) \, dx = x \sin(\log x) - \int x \cdot \cos(\log x) \cdot \frac{1}{x} \, dx = x \sin(\log x) - \int \cos(\log x) \, dx \]
Step 2: Conclusion.
Thus, the value of \( \int \sin(\log x) \, dx + \int \cos(\log x) \, dx \) simplifies to: \[ x \sin(\log x) + C \] Therefore, the correct answer is (3) \( x \sin(\log x) + C \).

Answer 2

Instead of integrating by parts directly, let's substitute \( t = \log x \), so \( x = e^t \) and \( dx = e^t\,dt \), turning both integrals into standard forms with known reduction formulas.

The two integrals become \( \int e^t \sin t \, dt \) and \( \int e^t \cos t \, dt \). Using the standard results \[ \int e^t \sin t\, dt = \frac{e^t(\sin t - \cos t)}{2} + C_1, \qquad \int e^t \cos t\, dt = \frac{e^t(\sin t + \cos t)}{2} + C_2, \] adding them gives \[ \frac{e^t(\sin t-\cos t)}{2} + \frac{e^t(\sin t+\cos t)}{2} = \frac{2e^t\sin t}{2} = e^t \sin t. \] Substituting back \( t = \log x \), \( e^t = x \), this becomes \( x\sin(\log x) + C \). Let's check this against each option.

1. \( \sin(\log x) + C \): This omits the factor of \( x \) that arises from \( e^t = x \) in the substitution; differentiating this would not reproduce \( \sin(\log x)+\cos(\log x) \), so it is incorrect.
2. \( \cos(\log x) + C \): Differentiating this gives \( -\sin(\log x)/x \), which does not match the integrand at all, so this is incorrect.
3. \( x\sin(\log x) + C \): This matches the result obtained from the substitution and reduction formulas above exactly.
4. \( x\cos(\log x) + C \): This has the right factor of \( x \) but the wrong trigonometric function; it corresponds to a different combination (such as \( \cos(\log x) - \sin(\log x) \) type integrands), not this one, so it is incorrect.

The substitution \( t=\log x \) combined with the standard \( e^t\sin t \), \( e^t \cos t \) reduction formulas confirms the sum of the two integrals.

Therefore, the correct answer is \( x\sin(\log x) + C \).