Question

Evaluate the integral: $$\int [\sin(\log x) + \cos(\log x)]\ dx$$

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

Hint:

You can also verify this instantly using the derivative product rule on the options!
Differentiating option (D) via the product rule:
$$\frac{d}{dx}[x\sin(\log x)] = 1 \cdot \sin(\log x) + x \cdot \cos(\log x) \cdot \frac{1}{x} = \sin(\log x) + \cos(\log x)$$ Since its derivative matches the integrand, option (D) must be the correct answer.

Answer 1

The Correct Option is D

Step 1: See the integral.
We must find $\displaystyle \int [\sin(\log x) + \cos(\log x)]\,dx$. The $\log x$ inside the trig functions is awkward, so we substitute.
Step 2: Substitute the log.
Let $\log x = t$, so $x = e^t$ and $dx = e^t\,dt$.
Step 3: Change the integral.
\[ I = \int (\sin t + \cos t)\,e^t\,dt \]
Step 4: Recall a standard result.
There is a neat rule: $\displaystyle \int e^t[f(t) + f'(t)]\,dt = e^t f(t) + c$. We try to match our integrand to this shape.
Step 5: Match the parts.
Let $f(t) = \sin t$. Then $f'(t) = \cos t$. So $\sin t + \cos t = f(t) + f'(t)$, which fits exactly. Hence \[ I = e^t \sin t + c \]
Step 6: Put $x$ back.
Since $e^t = x$ and $t = \log x$: \[ I = x\sin(\log x) + c \] \[ \boxed{x\sin(\log x) + c} \]