Question

Evaluate the following indefinite integral: \[ \int \sin(\log x)\,dx \]

• A. \(\dfrac{x}{2}[\sin(\log x)-\cos(\log x)] + C\)
• B. \(\dfrac{x}{2}[\cos(\log x)-\sin(\log x)] + C\)
• C. \(x[\sin(\log x)+\cos(\log x)] + C\)
• D. \(\dfrac{x}{2}[\sin(\log x)+\cos(\log x)] + C\)

Hint:

For integrals involving \(\log x\), try the substitution \(x=e^t\) or \(t=\log x\). This converts complicated logarithmic expressions into simpler exponential–trigonometric integrals.

Answer 1

The Correct Option is D

The given question asks us to evaluate the indefinite integral:

\(\int \sin(\log x)\,dx\)

To solve this integral, we will use integration by parts. Recall the integration by parts formula:

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

Here, we select:

• \(u = \sin(\log x)\)
• \(dv = dx\)

Then, we find:

• The derivative: \(du = \cos(\log x) \, \frac{1}{x} \, dx\)
• The integral of \(dv\): \(v = x\)

Applying the integration by parts formula, we get:

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

Simplifying further:

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

Notice, we now have a similar integral to solve. We use integration by parts again:

• \(u = \cos(\log x)\)
• \(dv = dx\)

Then, we find:

• The derivative: \(du = -\sin(\log x) \, \frac{1}{x} \, dx\)
• The integral of \(dv\): \(v = x\)

Applying the integration by parts formula again, we get:

\(\int \cos(\log x)\,dx = x \cos(\log x) + \int x \sin(\log x) \, \frac{1}{x} \, dx\)

Simplifying this gives:

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

We recognize that we've set up an equation in terms of the original integral:

\(I = x \sin(\log x) - [x \cos(\log x) + I]\)

Solving for \(I\) gives:

\(2I = x \sin(\log x) - x \cos(\log x)\)

Thus:

\(I = \frac{x}{2}[\sin(\log x) + \cos(\log x)] + C\)

The correct answer is:

\(\dfrac{x}{2}[\sin(\log x)+\cos(\log x)] + C\)