Question Statement

Evaluate the integral:

$\int \sin (\ln x)dx$

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Background and Explanation

This integral requires integration by parts, which reverses the product rule for differentiation. Since the integrand is a composition of trigonometric and logarithmic functions, we treat it as a product with $1$ and apply integration by parts twice. The key insight is that the second application returns us to the original integral, allowing us to solve for it algebraically.

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Solution

Let us denote the integral by $I$:

$I=\int \sin (\ln x)dx$

We rewrite the integrand as a product to prepare for integration by parts:

$I=\int \sin (\ln x)\cdot 1dx$

First application of integration by parts:

Choose:

• $u=\sin (\ln x)$, which gives $du=\cos (\ln x)\cdot \frac{1}{x}dx$ (by the chain rule)
• $dv=1dx$, which gives $v=x$

Using the formula $\int udv=uv-\int vdu$:

$\begin{array}{rl}I& =\sin (\ln x)\cdot x-\int x\cdot \cos (\ln x)\cdot \frac{1}{x}dx\\ & =x\sin (\ln x)-\int \cos (\ln x)dx\end{array}$

Second application of integration by parts:

Now we apply integration by parts to the remaining integral $\int \cos (\ln x)dx$. Again, we treat this as $\int \cos (\ln x)\cdot 1dx$.

Choose:

• $u=\cos (\ln x)$, which gives $du=-\sin (\ln x)\cdot \frac{1}{x}dx$
• $dv=1dx$, which gives $v=x$

Applying the formula:

$\begin{array}{rl}\int \cos (\ln x)dx& =\cos (\ln x)\cdot x-\int x\cdot \left(-\sin (\ln x)\right)\cdot \frac{1}{x}dx\\ & =x\cos (\ln x)+\int \sin (\ln x)dx\\ & =x\cos (\ln x)+I\end{array}$

Substituting back and solving for $I$:

Substitute this result back into our expression for $I$:

$\begin{array}{rl}I& =x\sin (\ln x)-\left[x\cos (\ln x)+I\right]\\ & =x\sin (\ln x)-x\cos (\ln x)-I\end{array}$

Now we have the original integral on both sides. Collecting terms:

$2I=x\sin (\ln x)-x\cos (\ln x)$

Factoring out $x$:

$2I=x\left[\sin (\ln x)-\cos (\ln x)\right]$

Solving for $I$:

$I=\frac{x}{2}\left[\sin (\ln x)-\cos (\ln x)\right]+C$

where $C$ is the constant of integration.

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Key Formulas or Methods Used

• Integration by parts: $\int udv=uv-\int vdu$
• Chain rule for differentiation:
  • $\frac{d}{dx}[\sin (\ln x)]=\cos (\ln x)\cdot \frac{1}{x}$
  • $\frac{d}{dx}[\cos (\ln x)]=-\sin (\ln x)\cdot \frac{1}{x}$
• Cyclic integral technique: When integration by parts returns the original integral, solve algebraically for the unknown integral

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Summary of Steps

1. Set up: Define $I=\int \sin (\ln x)dx$ and rewrite as $\int \sin (\ln x)\cdot 1dx$ to prepare for integration by parts.
2. First integration by parts: Choose $u=\sin (\ln x)$ and $dv=dx$ to obtain $I=x\sin (\ln x)-\int \cos (\ln x)dx$.
3. Second integration by parts: Apply the same method to $\int \cos (\ln x)dx$ with $u=\cos (\ln x)$, yielding $x\cos (\ln x)+I$.
4. Substitute: Replace the remaining integral to get $I=x\sin (\ln x)-x\cos (\ln x)-I$.
5. Solve algebraically: Rearrange to $2I=x[\sin (\ln x)-\cos (\ln x)]$.
6. Final answer: Divide by 2 and add the constant of integration: $\frac{x}{2}[\sin (\ln x)-\cos (\ln x)]+C$.