Question

The integral $\int \cos( \log \; x)dx$ is equal to : (where C is a constant of integration)

• A. $\frac{x}{2} \left[\sin\left(\log_{e} x\right)-\cos\left(\log_{e}x\right)+C\right]$
• B. $\frac{x}{2}\left[\cos\left(\log_{e}x\right) + \sin\left(\log_{e} x\right)\right]+C$
• C. $x [\cos (\log_e x) + \sin (\log_e x )] + C $
• D. $x [\cos (\log_e x) - \sin (\log_e x )] + C $

Answer 1

The Correct Option is B

To solve the integral $\int \cos( \log x) \, dx$, we can use a substitution method. Here's a step-by-step solution:

1. Let $u = \log x$. Therefore, the derivative of $u$ with respect to $x$ is $\frac{du}{dx} = \frac{1}{x}$. This implies $du = \frac{1}{x} \, dx$ or $dx = x \, du$.
2. Substitute $u$ and $dx$ in the integral: $$\int \cos(u) \cdot x \, du$$. Since $x = e^u$, we have: $$\int \cos(u) \cdot e^u \, du$$.
3. Apply integration by parts to solve this integral. Integration by parts formula is: $$\int u \, dv = uv - \int v \, du$$. Here, choose $v = e^u$ and $dw = \cos(u) \, du$: - $dv = e^u \, du$ - $w = \sin(u)$
4. Substitute into the integration by parts formula: $$\int \cos(u) \cdot e^u \, du = e^u \sin(u) - \int e^u \sin(u) \, du$$.
5. Apply integration by parts again to $\int e^u \sin(u) \, du$. Use: - $v = e^u$, $dw = \sin(u) \, du$ - $dv = e^u \, du$, $w = -\cos(u)$
   Then: $$\int e^u \sin(u) \, du = -e^u \cos(u) + \int e^u \cos(u) \, du$$.
6. This forms an equation: $$I = e^u \sin(u) + e^u \cos(u) - I$$, solving gives: $$2I = e^u \sin(u) + e^u \cos(u)$$, $$I = \frac{1}{2} e^u [\sin(u) + \cos(u)]$$. Reverting back to $x = e^u$, the result becomes: $$I = \frac{x}{2} [\sin(\log x) + \cos(\log x)]$$.
7. Therefore, the integral evaluates to: $$\frac{x}{2} \left[\cos(\log x) + \sin(\log x)\right] + C$$, where $C$ is the constant of integration.

Thus, the correct answer is: $\frac{x}{2}\left[\cos\left(\log_{e}x\right) + \sin\left(\log_{e} x\right)\right]+C$.