Question

\(\int \left[\sin(\log x)+\cos(\log x)\right] dx\) is equal to :

• A. \(-\cos(\log x)+C\)
• B. \(\frac{1}{\cos x}+\frac{1}{\sin x}+C\)
• C. \(\sin(\log x)+C\)
• D. \(x\sin(\log x)+C\)

Hint:

Put log x = t so that x = e^t and use the result that the integral of e^t (f + f prime) is e^t f.

Answer 1

The Correct Option is D

Step 1: Substitute to remove the logarithm:
Put \(\log x=t\). Then \(x=e^{t}\) and \(dx=e^{t}\,dt\).
\[ I=\int (\sin t+\cos t)\,e^{t}\,dt \]

Step 2: Recognise the standard form:
We know \(\int e^{t}\left[f(t)+f^{\prime}(t)\right]dt=e^{t}f(t)+C\).
Take \(f(t)=\sin t\). Then \(f^{\prime}(t)=\cos t\), so the integrand has exactly this form.
\[ I=e^{t}\sin t+C \]

Step 3: Go back to x:
Replace \(e^{t}\) by \(x\) and \(t\) by \(\log x\).
\[ I=x\sin(\log x)+C \]

Step 4: Why the other options are wrong:
(A) The derivative of \(-\cos(\log x)\) is \(\frac{\sin(\log x)}{x}\), not the integrand.
(B) The derivative of \(\frac{1}{\cos x}+\frac{1}{\sin x}\) has no logarithm in it.
(C) The derivative of \(\sin(\log x)\) is \(\frac{\cos(\log x)}{x}\), not the integrand.

Final Answer:
The integral is \(x\sin(\log x)+C\), option (D). \[ \boxed{x\sin(\log x)+C} \]