Question

Let 
\(I = \int_{e^{-\pi/2}}^{e^{\pi/2}} \left( \sin^2(\log(x)) + \sin(\log(x^2)) \right) \, dx\)
What is the value of $I$ ? 
 

• A. $e^{\pi/2} - e^{-\pi/2}$
• B. 0
• C. $\frac{\pi e^{\pi/2}}{2}$
• D. $e^{\pi} - 1$

Hint:

Whenever you see \(\log(x)\) in the integrand, substituting \(x = e^t\) often simplifies the integral into a form involving \(e^t\).
Always look for the standard form \(\int e^t(f(t) + f'(t)) dt = e^t f(t)\) to avoid tedious integration by parts.

Answer 1

The Correct Option is A

To solve the given integral \(I = \int_{e^{-\pi/2}}^{e^{\pi/2}} \left( \sin^2(\log(x)) + \sin(\log(x^2)) \right) \, dx\), we first need to simplify the expression inside the integral.

Let's break it down:

1. First, consider \(\sin(\log(x^2))\). Note that \(\log(x^2) = 2\log(x)\), so \(\sin(\log(x^2)) = \sin(2\log(x))\).
2. Using the trigonometric identity \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\), we can write: \(\sin(2\log(x)) = 2\sin(\log(x))\cos(\log(x))\).
3. Now combined, the integral becomes: \(I = \int_{e^{-\pi/2}}^{e^{\pi/2}} \left( \sin^2(\log(x)) + 2\sin(\log(x))\cos(\log(x)) \right) \, dx\)
4. Combine terms: \(\sin^2(\log(x)) + 2\sin(\log(x))\cos(\log(x)) = (\sin(\log(x)) + \cos(\log(x)))^2\)
5. Our integral simplifies to: \(I = \int_{e^{-\pi/2}}^{e^{\pi/2}} (\sin(\log(x)) + \cos(\log(x)))^2 \, dx\)

Next, we will use a substitution to simplify this integral:

1. Let \(u = \log(x)\). Then \(du = \frac{1}{x} \, dx\), so \(dx = x \, du\). Also note that as \(x\) goes from \(e^{-\pi/2}\) to \(e^{\pi/2}\), \(u\) goes from \(-\pi/2\) to \(\pi/2\).
2. Rewrite the integral in terms of \(u\): \(I = \int_{-\pi/2}^{\pi/2} (\sin(u) + \cos(u))^2 e^u \, du\)
3. Expand and simplify: \((\sin(u) + \cos(u))^2 = \sin^2(u) + 2\sin(u)\cos(u) + \cos^2(u)\) Using \(\sin^2(u) + \cos^2(u) = 1\), this becomes: \(1 + 2\sin(u)\cos(u)\).
4. So the integral becomes: \(I = \int_{-\pi/2}^{\pi/2} \left(1 + 2\sin(u)\cos(u)\right)e^u \, du\)
5. Break this integral into parts: \(I = \int_{-\pi/2}^{\pi/2} e^u \, du + 2\int_{-\pi/2}^{\pi/2} \sin(u)\cos(u) e^u \, du\)

Evaluate each part separately:

1. The first integral: \(\int_{-\pi/2}^{\pi/2} e^u \, du = \left[ e^u \right]_{-\pi/2}^{\pi/2} = e^{\pi/2} - e^{-\pi/2}\)
2. The second integral: \(\int_{-\pi/2}^{\pi/2} \sin(u)\cos(u) e^u \, du = 0\) due to the symmetry of \(\sin(u)\cos(u)\) around \(u = 0\)

Thus, combining the results:

\(I = e^{\pi/2} - e^{-\pi/2} + 2 \times 0 = e^{\pi/2} - e^{-\pi/2}\)

Therefore, the value of \(I\) is \(e^{\pi/2} - e^{-\pi/2}\).