Question

\( \int e^x(x^2-2)\cos(e^x(x^2-2x)) dx = \)

• A. \( \sin(e^x(x^2-2x)) + C \)
• B. \( \sin(e^x(x^2-2)) + C \)
• C. \( x^2e^x\sin(e^x(x^2-2)) + C \)
• D. \( e^x\sin(e^x(x^2-2)) + C \)
• E. \( e^x\sin(x^2e^x-2xe^x) + C \)

Hint:

If integrand is of form \( f'(x)\cos(f(x)) \), answer is \( \sin(f(x)) \).

Answer 1

The Correct Option is A

Note: There seems to be a significant typo in the question's integrand versus the argument of the sine function in the answer. Let's analyze. The integral is \(\int e^x(x^2-2)\cos(e^x(x^2-2x)) dx\). The answer is \(\sin(e^x(x^2-2x)) + C\). Let's check if the derivative of the answer gives the integrand. Let \(u = e^x(x^2-2x)\). The derivative of \(\sin(u)\) is \(\cos(u) \cdot u'\). We need to find \(u'\): \(u' = \frac{d}{dx}[e^x(x^2-2x)]\). Using the product rule: \(u' = (e^x)(x^2-2x) + e^x(2x-2) = e^x(x^2-2x+2x-2) = e^x(x^2-2)\). So, the derivative of \(\sin(e^x(x^2-2x))\) is \(\cos(e^x(x^2-2x)) \cdot e^x(x^2-2)\). This perfectly matches the integrand. Therefore, the question had a typo and the argument of cosine should be \(e^x(x^2-2x)\). Let's solve with the corrected version. Step 1: Understanding the Concept:
This integral is solvable by u-substitution. The structure of the integrand is \(\cos(g(x)) \cdot g'(x)\), which is the result of applying the chain rule to \(\sin(g(x))\).
Step 2: Key Formula or Approach:
1. Let \(u\) be the inner function inside the cosine: \(u = e^x(x^2 - 2x)\). 2. Calculate the differential \(du\). 3. Check if \(du\) matches the remaining part of the integrand. 4. Substitute to get an integral of the form \(\int \cos(u) du\). 5. Integrate and substitute back.
Step 3: Detailed Explanation:
The integral is (with correction): \[ \int e^x(x^2-2) \cos(e^x(x^2-2x)) dx \] Let's make the substitution: \[ u = e^x(x^2 - 2x) \] Now, let's find \(du\) by differentiating `u` with respect to `x` using the product rule: \[ \frac{du}{dx} = \frac{d}{dx}(e^x) \cdot (x^2 - 2x) + e^x \cdot \frac{d}{dx}(x^2 - 2x) \] \[ \frac{du}{dx} = e^x(x^2 - 2x) + e^x(2x - 2) \] \[ \frac{du}{dx} = e^x(x^2 - 2x + 2x - 2) = e^x(x^2 - 2) \] So, the differential is: \[ du = e^x(x^2 - 2) dx \] This exactly matches the part of the integrand outside the cosine function. Now, we can substitute `u` and `du` into the integral: \[ \int \cos(u) du \] The integral of \(\cos(u)\) is \(\sin(u)\): \[ \sin(u) + C \] Finally, substitute back \(u = e^x(x^2-2x)\): \[ \sin(e^x(x^2-2x)) + C \] Step 4: Final Answer:
The integral is \(\sin(e^x(x^2-2x)) + C\).