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

•

Step 1 : Understanding the Question:
We are given a definite integral involving trigonometric functions of logarithmic arguments.
We need to evaluate the value of the integral \(I\). between the limits \(e^{-\pi/2}\) and \(e^{\pi/2}\).

•

Step 2 : Key Formula or Approach:
First, we simplify the integral using substitution.
Let \(t = \log(x)\)., which implies \(x = e^t\) and \(dx = e^t dt\).
Also, note that \(\log(x^2) = 2\log(x) = 2t\).
This transforms the integral into the standard form:
\[ \int e^t \left( f(t) + f'(t) \right) dt = e^t f(t) + C \] We identify \(f(t) = \sin^2(t)\) and verify if \(f'(t) = \sin(2t)\).

•

Step 3 : Detailed Explanation:
Let us apply the substitution:
\[ t = \log(x) \implies x = e^t \implies dx = e^t dt \] Let us determine the new limits of integration:
When \(x = e^{-\pi/2}\)., we have \(t = \log(e^{-\pi/2}) = -\frac{\pi}{2}\).
When \(x = e^{\pi/2}\)., we have \(t = \log(e^{\pi/2}) = \frac{\pi}{2}\).
Now, rewrite the integrand:
\[ \sin^2(\log(x)) = \sin^2(t) \] \[ \sin(\log(x^2)) = \sin(2t) \] Substituting these into the integral:
\[ I = \int_{-\pi/2}^{\pi/2} \left( \sin^2(t) + \sin(2t) \right) e^t dt \] Let us define \(f(t) = \sin^2(t)\).
Its derivative is:
\[ f'(t) = \frac{d}{dt} \left( \sin^2(t) \right) = 2\sin(t)\cos(t) = \sin(2t) \] Thus, the integrand is exactly of the form \(e^t (f(t) + f'(t))\).
The antiderivative is:
\[ F(t) = e^t \sin^2(t) \] Now we evaluate the definite integral using the limits:
\[ I = F\left(\frac{\pi}{2}\right) - F\left(-\frac{\pi}{2}\right) \] \[ I = e^{\pi/2} \sin^2\left(\frac{\pi}{2}\right) - e^{-\pi/2} \sin^2\left(-\frac{\pi}{2}\right) \] Since \(\sin\left(\frac{\pi}{2}\right) = 1\) and \(\sin\left(-\frac{\pi}{2}\right) = -1\)., their squares are both 1:
\[ \sin^2\left(\frac{\pi}{2}\right) = 1, \quad \sin^2\left(-\frac{\pi}{2}\right) = 1 \] Therefore:
\[ I = e^{\pi/2}(1) - e^{-\pi/2}(1) = e^{\pi/2} - e^{-\pi/2} \]
•

Step 4 : Final Answer:
The value of the integral is \(e^{\pi/2} - e^{-\pi/2}\).
This corresponds to option (A).