Calculate the integral: integral from 1 to e (sin(ln

Calculate the integral: $\int_1^e \frac{\sin(\ln x)}{x} \, dx$.

We will use substitution to transform this integral into a simpler form.

Step 1: Choose a substitution

Let $u = \ln x$. Then $du = \frac{1}{x} \, dx$, which means $\frac{dx}{x} = du$.

Step 2: Change the limits of integration

When $x = 1$: $u = \ln(1) = 0$. When $x = e$: $u = \ln(e) = 1$.

Step 3: Rewrite the integral

\[\int_1^e \frac{\sin(\ln x)}{x} \, dx = \int_0^1 \sin(u) \, du\]

Step 4: Find the antiderivative

We know that:

\[\int \sin(u) \, du = -\cos(u) + C\]

Step 5: Apply the Newton-Leibniz formula

\[\int_0^1 \sin(u) \, du = \left[-\cos(u)\right]_0^1\]

Step 6: Evaluate at the limits

At $u=1$: $-\cos(1)$. At $u=0$: $-\cos(0)=-1$.

Step 7: Calculate the definite integral

\[\int_0^1 \sin(u) \, du = -\cos(1) - (-1) = 1 - \cos(1)\]

Numerical value:

Using $\cos(1) \approx 0.5403$ (radians):

\[1 - \cos(1) \approx 1 - 0.5403 = 0.4597\]

Alternative verification:

We can verify that our substitution was correct by checking the derivative: if $F(x) = -\cos(\ln x)$, then

\[F'(x) = -\frac{d}{dx}[\cos(\ln x)] = -(-\sin(\ln x)) \cdot \frac{1}{x} = \frac{\sin(\ln x)}{x}\]

This confirms that $-\cos(\ln x)$ is indeed an antiderivative of $\frac{\sin(\ln x)}{x}$. Using this antiderivative directly:

\[\int_1^e \frac{\sin(\ln x)}{x} \, dx = \left[-\cos(\ln x)\right]_1^e = -\cos(\ln e) - (-\cos(\ln 1)) = -\cos(1) + 1 = 1 - \cos(1)\]

This matches our result!

Geometric interpretation:

The integral $\int_1^e \frac{\sin(\ln x)}{x} \, dx$ represents the area under the curve $y = \frac{\sin(\ln x)}{x}$ from $x = 1$ to $x = e$. At $x=1$: $y=\frac{\sin(0)}{1}=0$ (the curve starts at the $x$-axis); at $x=e$: $y=\frac{\sin(1)}{e}\approx 0.3096$. The function is positive throughout $[1,e]$ since $\ln x \in [0,1]$ and $\sin(u)>0$ for $u\in(0,1)$. Through the substitution $u=\ln x$, we transformed this into the simpler problem of finding the area under $y=\sin(u)$ from $u=0$ to $u=1$.

Connection to logarithmic functions:

This integral demonstrates a common technique: when you see $\frac{f(\ln x)}{x}$, the substitution $u = \ln x$ (so $du = \frac{dx}{x}$) often simplifies the problem significantly.

Final answer:

\[\boxed{\int_1^e \frac{\sin(\ln x)}{x} \, dx = 1 - \cos(1)}\]

Or numerically:

\[\boxed{\int_1^e \frac{\sin(\ln x)}{x} \, dx \approx 0.4597}\]