Question:
\( \int e^x(x^2-2)\cos(e^x(x^2-2x)) dx = \)
\( \int e^x(x^2-2)\cos(e^x(x^2-2x)) dx = \)
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If integrand is of form \( f'(x)\cos(f(x)) \), answer is \( \sin(f(x)) \).
Updated On: Apr 21, 2026
- \( \sin(e^x(x^2-2x)) + C \)
- \( \sin(e^x(x^2-2)) + C \)
- \( x^2e^x\sin(e^x(x^2-2)) + C \)
- \( e^x\sin(e^x(x^2-2)) + C \)
- \( e^x\sin(x^2e^x-2xe^x) + C \)
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The Correct Option is A
Solution and Explanation
Concept:
Recognize derivative inside cosine.
Step 1: Let \( u = e^x(x^2-2x) \).
Step 2: Then \( du = e^x(x^2-2x) + e^x(2x-2) = e^x(x^2-2)\,dx \).
Step 3: Substitute.
\[ \int \cos(u)\,du = \sin(u) + C \] \[ = \sin(e^x(x^2-2x)) + C \]
Step 1: Let \( u = e^x(x^2-2x) \).
Step 2: Then \( du = e^x(x^2-2x) + e^x(2x-2) = e^x(x^2-2)\,dx \).
Step 3: Substitute.
\[ \int \cos(u)\,du = \sin(u) + C \] \[ = \sin(e^x(x^2-2x)) + C \]
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